 Last class, you basically discussed about operations on complex numbers. In fact, I talked about representation of complex numbers, point form, polar form, Euler's form, and we also did basic operations like comparison, addition, subtraction, multiplication, division. We talked about conjugate. We talked about square rooting. We talked about logarithm. Okay. So in all these operations, there is one operation which is left off, which is raising any power, any power of integer or a rational number on an integer on a complex sum up. So we are going to start with that concept. And that concept is called as the demoverase theorem. D demoverase, let me just, D demoverase theorem. Was this theorem taken up in school when you did complex numbers? Did you come across demoverase theorem? The pronunciation is more of a French word. Okay. So in this, we are going to learn how to raise a complex number. Right. I actually remember it by that name only. So I'm telling you my inside method to remember this name. So it's D mohave. So let's say you have a complex number, which is a given by R says theta. So I've chosen a polar form. And if I'm raising it to the power of N, okay, if I raise it to the power of N, what happens? Now, please do not restrict yourself to thinking that N could only be natural numbers, right? N could be negative integers also and could be a rational number of P by Q form as well. So I'm going to divide this theory into two parts. One is where I'm going to talk about case number one, where your N happens to be an integer. Okay. Where your N is an integer and case two is where I'll be talking about N being a rational number of the nature. P by Q. Okay. Now, for an integer, the concept is pretty simple. If you have a complex number Z, which is raised to the power of an integer, then what happens? Then what happens? You raise the modulus to the same integral power. Okay. And you multiply the argument by that integer. Okay. So as you can see here, R is raised to the power of N and the argument has been multiplied with N, right? Now, this is very evident actually because if you look at the Euler's form notation, you can easily understand how this or why this works actually. So if you write the thing in Euler form R e to the power i theta and raise it to the power N, it is as good as R to the power N e to the power i N theta. So now this becomes your new modulus and this becomes your new argument. So after raising it to a power of N, your modulus gets raised to the same power of N, whereas the argument gets multiplied to that N. Okay. A simple example I can cite here. Let's say I have a complex number. I just take a simple case two sys pi by six. Okay. Two sys pi by six. And somebody says, I want to raise this complex number to the power of five. Okay. So what should be the result? So simple as per demoverage theorem, you don't have to sit and multiply them to the power, multiply them five times or apply any kind of a binomial expansion. Demoverage theorem basically makes your life pretty easy. The answer to the power five will be two to the power five into cos five five by six plus I sign five five by six. Of course, I can further simplify it. Okay. And if you want to convert it to point form, you can always do that, but I'm not interested in simplifying it to any point form. Just wanted to demonstrate how this particular thing works. Okay. So, remember when you raise a complex number to the power of N. Okay. This gives you only one answer and this is the answer. So you get only one answer. Only one answer out of it. Okay. Only one answer. Why I'm saying this. Yeah. Only one answer. Why I'm saying this is because when I take N as a fraction, maybe a rational number, I will get multiple answers from there. We'll talk about it when the right time comes. Okay. As of now, let us understand this first. So please note down people who have joined in late. First of all, good afternoon. And we were doing operations on complex numbers where I am basically talking about discussing how to raise a power on a complex number, which is a rational number, but rational number basically at first taken it as integer first. So for integer, basically the rule works like this. The modulus gets raised to the same integer power, and the argument gets multiplied to that integer. Okay, I give one demonstration for the same. Okay. The few points to be noted here, please note down the following points. Number one. If your complex number is given to you like this R cos theta plus I sign phi. Where theta and phi are not equal to each other. Okay. And somebody says, hey, I want to raise this power. I want to raise this complex number to integer power and okay. Please note this down, you cannot write it like this. That means the demorbid theorem is not going to work on this. Okay, please note. This is a z to the power n will not be equal to R to the power n cos n theta plus I sign and fight because primarily this is not the polar form for it as I already told you in polar form theta and phi are equal to each other. So this is not the polar form. Hence this result will not come out. In many competitive exams, they will like they will try to deceive you. They will try to deceive you with a similar looking expression to that of a polar form and they will ask you that if I raise this to the power of n, which of the following options can you know occur and basically one of the options will be this so please do not market please do not market. This is not going to work out. Okay. The second imposter which you should be aware of. If I write my complex number in a slightly fancy way. Let's say sine theta plus I cos theta. Okay. And I want to raise z to the integer power and then also this is not going to be R to the power and sine and theta plus I cos and theta so please note these two are not going to work out. Okay. So be aware, be careful. I'm just, you know, giving you some kind of imposter imposter means they will try to deceive you but don't get deceived but don't get deceived. So this was the story related to the first part of the Maurese theorem when our power was an integer. Now integer could be both positive or negative so don't be under the impression that Saris saying integer means he means natural number. No. The same rule could be applied even for a negative integer also. Okay. The second rule is slightly tricky and lengthy which I will be now giving in some time but if you want to note down anything from this page please do so. Please do so. Could you scroll up. Okay. Power zero will become one in this case also yeah yeah power zero will become one in this case also where you want me to stop Manu, do you want me to scroll all the way to the first line that I wrote or somewhere in between. I mean, is this position okay. Okay fine. Done. Okay, thank you. I will be moving towards the second case where your N is not an integer rather rather N is a rational number. Okay. That is N is of the form P by Q where P and Q are integers where P and Q are integers. Okay. Now to be going into the proof of this I'll be directly stating the result because proof is not going to be required in any form or shape either in J main or in J advance. Okay. Maybe when you do complex number analysis in your undergraduate, there the proof might be you know stated to you. The form of the system gives a very, very important interesting in a rule for finding out a power of P by Q raised to a complex number. So let's say I write the complex number as R says theta. Okay, and I want to raise this complex number to the power of P by Q. Okay. So basically this is what I am interested in doing. Okay. Now let us try to understand this expression slightly more deeply. When you raise a number to the power of P by Q, let's see this in phases. It is as good as first you're raising it to the power of P. Okay. And by the way, when I write this P by Q is in the simplest form P by Q has been expressed in the simplest form expressed in the simplest form simplest form is they cannot be further scope of canceling out anything. Okay. P by Q has been expressed in their simplest form. Okay. Now, raising it to the power of P by Q, I'm just carrying out this operation in two phases. One is I'm raising it to the power of P. And then I'm raising this whole thing to the power of one by Q. Okay. Now P being an integer, if you see the inside part will follow the demoverage theorem which we had discussed in case one where our power was integer. So that will become R to the power P cos P theta plus I sign P theta. Okay. I think nobody has any doubt or concerns till this stage because this is very much what we did in the previous slide. Okay. Is it fine any questions here. Now, now understand this, when you're raising a number to the power of one by Q, you are actually finding the Q with roots, just like square roots. When you're, when you're raising it to the power of half, you're finding it square roots in square roots part, which we did in the last class, we got two answers isn't it to square roots. Now, apply the same logic here and tell me how many answers will come out if I do this to the power of one by Q. That means I'm finding cute roots. And how many answers will come from it. Q answers will come right so we will end up getting Q answers from it. In the previous demoverage theorem form where we are where power was just an integer, we only got one answer right. R to the power n cos n theta plus I sign n theta or R to the power n sis n theta. But unlike the previous case here I will get two different answers. Right. So basically the denominator of your power that many number of roots I will get. Are you getting my point or that many number of answers I will get. So how do you find these Q answers so for that demoverage gave a very interesting mechanism. He said that raise this raise this R to the power p to the power of one by Q. Of course, when we when you're doing it keep it as a real. Okay, so this term will be completely real. So what is the whatever is the real answer that comes out from it. That answer you have to keep away because ultimately this is going to play the role of the modulus modulus is a real quantity because it signifies the distance of that complex number from the origin right. So then he said, and then he said, write this p theta as 2k pi plus p theta, like this. Okay. Now please understand here, when you change your p theta with 2k pi plus theta k being some integer I will tell you what integers k can take. Please note that both cost and sign are periodic with two pi. Correct. So changing your angle with a multiple of two pi is not going to impact the result that means cost p theta and cost 2k pi plus p theta would still be the same values. Isn't it, isn't it cost 30 degree and cost 2 pi plus 30 degree or cost 4 pi plus 30 degree or cost 6 pi plus 30 degree or cost minus 2 pi plus 30 d doesn't make any difference to your answer answer is still cost 30 root 3 by 2. The same goes with sign as well. Sign is also periodic with 2 pi. So any change in the angle by a multiple of 2 pi is not going to affect the trigonometric ratio. Okay. But this guy said, okay, after you have done this, take this one by q and multiply it with the arguments which you see over here. In short, okay, do one by q multiplication with the argument, right, just like we used to do in the case one of our demo best, you know, okay. And then he said, give k give k any q consecutive integral values. Normally we give it as 123 till q minus one. Okay. But in reality, you can actually give any q consecutive integral values. Integral values. As you provide k, these q consecutive integral values, one one root will come out from there. So when you put k consecutive, sorry, q consecutive integral values, you will end up getting q different answers, and those answers would be the answer for z to the power p by q. Okay, now I'm not going to give you the proof for this. We don't require the proof for this, right, you just have to understand this mechanism. Okay, so he said that all your roots, or all your answers of z to the power n, which is p by q in this case, can be obtained from here, which is r to the power p by q cost to k pi plus p theta by q whole divided by q. Plus I sign to k pi plus p theta whole divided by q where k can take any q consecutive integral values, but preferably start with zero, because they are I mean, many people ask me, sir, if I go beyond it, what will happen if I know if I put case q, what will happen if you put case q, you will end up getting the same answer what you got for zero. If you put q plus one, you'll get the same answer as what you got for one. So there's no point going beyond q minus one, because there will be a repetition of your roots happening. Once again the whole process. What do we do, what we can do say to you, we can take an example example will help us understand this. Okay, yeah, yeah, we'll take an example don't worry. So keep this result in your mind on your notes, we'll discuss it through an example. Okay, so let's say I want to. Let's say I take a complex number one. Okay, one is also complex number isn't it. Okay, I've taken a simplest case. Now I raise this complex number to the power of p by q where p is one and q is three. That means I'm finding the cube root of one. I had actually taken this case very briefly in the bridge course, where I told you that if you raise one to the power of one third you'll get three answers. What we actually knew that time was only one of the answers which was one. Okay, so that is only one of the answers, there are two more actually. So today is the right opportunity that we find what are the other two values. Okay, and we'll talk about it also because it's a part of our syllabus for J. So this is a complex number if you write it, it's one cost zero plus I sign zero. Okay. Thankfully, P is actually a one so I don't have to worry too much about multiplying with one because it will not affect our answer. So it directly becomes something like this cost zero plus I sign zero to the power of one third. Okay, now people who joined in late or you know for that matter say to as well. Say to what you do next is you you change your for sorry one to the power one by three can I just write it as a one I don't want to complicate it like writing it as one to the power one by three. You just write your angle here as 2k pi plus zero. Okay, basically I'm at this step now. I'm at this step now. Okay, okay, there was a power of one by q I was at that step which I've erased it now multiply it with one by q that means multiply this with one third. That means you have to divide this by three. So right now you are at this step, this step here. Okay, understood. Now, k has to be provided with values starting from, I mean you can provide any three consecutive integer values but we normally start from zero. Normally start from zero it is not a compulsion. So you can put zero one and go up till q minus one see remember here q is q is three, so you have to go till q minus one which is two. So now what will happen if you put k as zero let's check. If you put k as zero you'll end up getting cost zero by three I sign zero by three. What is that. What is cost zero plus I sign zero one only correct. So this is one of the real roots which we already know I mean even without solving it we knew that one of the answers would have been one only. If you put k as one, it'll give you cost two pi by three plus I sign two pi by three. By the way, if you simplify this this gives you minus half plus I root three by two. If you put k as a two, if you put k as a two, your answer will be cost four pi by three plus I sign four pi by three. Okay, and if you simplify this in fact this is not a polar form for it. The polar form for it is actually cost minus two pi by three plus I sign minus two pi by three. Why this is not a polar form because this angle is not in the principle argument right you should normally keep it as a principle argument. Anyways, these two whatever if you calculate it gives you minus half minus I root three pi. Okay, so these are the three cube roots of unity. If you cube each one of them you will see that you will always get a one as your answer. Try cubing this maybe after the class is over you can try it out try cubing this you will get a one of course one cube is and he was a one. Okay. Is the process understood now. Clear. Okay, now I'll give you a small question to work on. Everybody, please find out. Okay, find cube roots of minus one. Do it and let me know by writing it done on the chat box that you're done. Are you clear with the approach now. Any. Yeah, any discomfort with respect to any step do let me know okay. Yes, that is one of the answers what are the other two answers. One, when we want to find some of roots is it helpful to take k plus one from minus k by two to k by two or sums some of the roots you want some of the roots will be zero always. You're talking about these cases one to the power one by three or something. Yeah, some of the rules will always be zero. Now, very interestingly why it will be zero I'll discuss about that also because it's there in my agenda. See, when you have these kind of complex numbers that's a one z to the power one third, and you're trying to find out the complex roots of it okay. So basically you're trying to find out. Let's say I take a variable x. Okay, let's say x is one of the roots of it. That means you're trying to find out the roots of this equation right and in this equation, you will see that there is no x square component. There's no x square term right. And you would have already done it in class 10 that the sum of the roots of a cubic equation is minus B by a. Right, it's actually true for any polynomial equation. So if you see there is no B term error B zero. So B by one, that is zero by one will anyways be a zero. Okay, so some of the roots here will definitely be zero. I will talk about it this is there in my discussion list today. Yes, so are you all done with this. If you don't mind, can you type your answer on the chat box if you want to. We'll discuss everything we'll discuss everything. Okay, so first of all, let's write down one as a polar form. So one as a polar form is our sis pi isn't it. I'm raising it to the power of one third that means I'm raising this to the power of one third. So as already discussed, this will become cost to k pi plus pi by three. I sign to k pi plus pi by three. Okay, and you have to, you have to put your k values as consecutive three values, you can see two zero one and two. Okay, so when you put a zero. When you put a zero, the complex number that you attain is cost pi by three plus I sign pi by three. By the way, when you simplify it, it gives you half plus I root three by two. Okay. And if you put a one, it'll give you, it'll give you costs. This will become a pie. It'll also become a pie and the answer will be minus one. So minus one definitely will be one of my answers, which I think that age already told. Okay. And the third answer that I will get will be when you put causes to so if you put causes to you'll get cost five pi by three. Okay, plus I sign five pi by three cost five pi by three is a half and sign I pi by three is a negative root three by two. Okay, so these are your three cube roots of negative one. These are your three cube roots of negative one. Okay. Now going back, going back to the square roots. So yeah, cube roots of one. Please note down this particular complex number is famously written as omega. Now why famously written as omega. Many times you will see that in your questions, they will say, there's a complex number omega, which is a complex cube root of unity. That means they're referring to this particular number as the as omega. Okay. But please note omega in general doesn't represent minus half plus I root three by two omega could be used for any complex number also. Normally Z omega, alpha, etc. We keep on using for complex number also. So let's say they say there is a complex number omega and they further, you know, give some other details to you. Please do not assume that that omega is actually minus half plus I root three by two. So please watch out for the phrase that omega is a complex cube root of unity. Okay, so this is what they call as call as omega. Okay. And this term, which is basically the second term that you see over here. This is omega square. This is omega square. Now, is omega square a kind of a symbol as it actually the square of omega. Yes, it is actually the square of omega, because if you square this up. You will realize that by demo of a theorem will give you cause four pi by three plus I sign four pi by three, which is exactly this complex number. So this is called omega and this is called omega square. So hence we say the cube roots of unity we normally write cube roots of unity or cubes of one we normally write them as one omega and omega square. Now you must be thinking said they look like to be in a geometric progression. Yes, they are in a geometric progression. Okay. It's just a name given to it. Why do you call, why do you call angular velocity as omega? Why do you call your angular acceleration as alpha just a name given to it people refer to it. That's it. No story behind it. No story behind it. Okay. Is it fine. Any questions. So, I gave you an exercise of finding the cube roots of negative one, you can note down here that the cube roots of negative one. This term is actually a negative omega negative omega square and this term is actually negative omega. Okay. So note down this is something that you can remember also cube roots of cube roots of negative one are minus one minus omega and minus omega square. So if you know cube roots of unity, you can also find out cube roots of minus one. Okay. Now we will analyze, you know, this particular rule. Through some problems. So let us look at look into some problems where we will apply our both the formats of demoverase theorem. Okay. And we will try to understand more through the problem solving aspect. Okay, anything that you want to copy here or anything that you would like to ask her please do so. These three are in geometric progression. What do you think are they in the geometric progression. Yes, they are common ratios omega. Okay, I will come back to cube roots of unity but only after taking one or two questions. Okay, so let me take some questions. Okay, so let me begin with a simple question here. So this cos theta plus I sin theta to the power four divided by sin theta plus I cos theta to the power five express this as a plus IB. In short, give the value of a and b. Of course it'll come out in terms of cos and sin of theta and all. I mean, theta would definitely be there in some form. So give me the value of a and b. Done. Can you tell what is your a value coming out and be value coming out. See, there's no problem with the numerator numerator is plain and simple it is very much in the in the form where we can easily apply our demoverage theorem. Okay, the denominator is not in the form which we would like to see it in so what I'll do is I will slightly focus on the denominator aspect of it. First of all, this term itself if I take a common. Okay, if I take an I common I'll end up getting cos theta minus I sin theta correct me if I'm wrong. Okay, now treat this treat this as cos minus theta plus I sin minus theta. Now opposite of demoverage theorem if you apply it is actually cos theta plus I sin theta to the power of minus one isn't it minus one only on this okay not on the entire thing. Now, the given expression to me is sin theta plus I cos theta and if I raise it to the power of five, it is as good as I to the power of five. And this is nothing but cos theta plus I sin theta to the power of minus five. I to the power of five is I correct. And let's not disturb this expression because we can simplify our complete expression by using this format only. Now let us come back to our question. Okay, let's look at this expression. Can I write this expression now as cos theta plus I sin theta to the power of four. The denominator I have reduced it to I times cos theta plus I sin theta to the power of minus five. Okay, you may use your exponent laws here and write it as cos theta plus I sin theta to the power of nine. One by I is minus I please note that one by I is a minus I now as per demoverage theorem I can write this as cos nine theta plus I sin nine theta. Once you multiply your minus I throughout you'll end up getting sine nine theta minus I cos nine theta. And just compare it with a plus IB compare it with a plus IB. So when you do that a becomes sine nine theta and B becomes minus of cos nine theta. I hope those are solid you would have got this as your answer. What is four theta minus five theta that also you will not simplify. But still, have you seen people giving answer like two plus three. Write a five. Okay, how many of you got this answer type of me on the chat box. Nobody, what is happening. Okay, we'll take another problem. Any question you have here please do ask any question that you have please do ask. Fine. So my next question is my next question is, find, find minus 32 to the power one by four, that means find four fourth roots of minus 32. Find for fourth roots of minus 32. Sorry, not minus 32 my bad. Let's keep it 16 minus 16. Just to keep it simple. Okay, minus 16. Okay, fine. So four fourth roots of minus 16. Okay. That's a first part of the question second part of the question is, locate the roots, locate the roots on the argon pain. That's it. Please solve this question. Find minus 16 to the power of one by four. Locate the, the four fourth roots of minus 16 on the argon pain. Once you've located will discuss a little bit more about it. Just follow the demoverious theorem that we had discussed in the beginning of our today's class. Okay, just follow step by step. Meanwhile, when you solve that I will give you a shorter way to find all the roots. Okay. You don't have to follow every step of demoverious theorem we can cut corners and we can solve these questions in a faster way, which will be very helpful for you in your examination scenario when you have lack of time. Should I also participate along with you. So 16 minus 16 as a polar form is our sis pi. Isn't it. R cos pi plus. I signed by correct. Now you're raising it to the power of one fourth. So basically this is what you all, you will be getting. Okay, correct me if I'm wrong. I'm just skipping some steps, because I don't want to write each and every step over here. I sign. Okay, so it's a very good attempt. We'll check it out whether your answer is correct or not. Now 16 to the power one by four, this itself will give you four answers. Right. So here is the thing that I was actually talking about in the, you know, when I was discussing the demoverious theorem in the beginning of today's session that you have to maintain a real positive value over here. A real positive value you have to keep. So which is the, you know, direct answer that comes in your mind when you do fourth root of 16, which is a real positive answer that comes in your mind. See, if you just talk about real, there could be two answers, four and, sorry, two and a minus two, but minus two cannot be your modulus right modulus of a complex number is always a positive quality. So think of a real positive answer that comes in your mind when you do fourth root of 16 the answer only answer is a two in your mind right. Okay, so keep right it as a two no need to, no need to you can just take a pi common here also if you want. Now start putting values of K. You could put any four consecutive integral values, but preferably we start with zero only so zero. One, two, and let me make some space for myself. Okay, when you put a zero, the very first answer that you will get as your one of the fourth roots will be two cos pi by four plus I sign pi by four. Two cos pi by four plus I sign pi by four correct me if I'm wrong. Clear any questions. Okay. The next route that you will get when you put K as a one is two cos three pi by four plus I sign three pi by four right now I'm just writing the general argument okay. I'm not converting it to the principal argument. Okay, so whatever answer comes out from putting K value that I'm writing as such I'm not making any kind of a change to the general argument, because I want to show you some analysis. Now if you put case two you get cos five pi by four plus I sign five pi by four. Okay. And when you put case three you get cos seven pi by four plus I sign seven pi by four. Is it fine. So far so good. Any questions. Okay. Okay. So if you locate these four roots of minus 16 on the argon thing, they will be located at these positions. All of you please pay attention. The first route will be located at a distance of two and at an argument of pi by four. Okay, so at a distance of two and at an argument of pi by four the first route will be located. So I'll be just making a simple name diagram over here. And I'll locate my first route. Okay. Can I call them as alpha beta gamma delta just to refer to them. Alpha beta gamma delta. So this is your alpha position. Okay. So this complex number of this route will be your alpha. Okay. Second route if you see carefully, that is also located at the very same distance from the origin which is to because it's found this is also to. But if you see that argument has increased by argument has increased by pi by two. Correct. Pi by two is actually two pi by four. Now why I'm calling it as two by four, two pi by four because I'll be showing you a trick of a simple shortcut way. So at a difference of pi by two from the alpha position. Okay. And at a distance of two you will get your second route. Let's call it as beta. So please note that this gap will be this gap will be 90 degree. And you'll see that the same gap of 90 degree will be continuing for gamma as well from beta. So gamma will also be at a gap of 90 degree. Okay. And same at a distance of two only just like these complex where numbers where from the origin. So this will be a gamma and delta would be again at a 90 degree 90 degree. Phase difference or argument difference you can say. Okay. So this will also be this will also be 90 degrees. And of course this will also be 90 degrees only. This will be a gamma. Okay. Now looking at this looking at this entire scenario. I have got a simple technique to look at these routes without actually doing all these stuffs. Okay. Now what is the technique? I'll come back. I'll come to that technique now here. Okay. So please listen to this trick to find any complex number to the power of P by Q. Okay. Take to find any complex number to the power of P by Q. Okay. So all of you please listen to this very, very important. Okay. You have to do the following steps. First step is first step is find by the way you are only given this complex number. You will be always already be provided with this complex number. Don't worry about it. So you would be knowing that you will be knowing the modulus and you would be knowing the argument of that. Right. So first step is find first step is fine. R to the power of P by Q. Okay. Please take the real positive number because it represents the modulus of your roots. Modulus is always a positive real quantity. Okay. Right. First step is find R to the power P by Q. Okay. Take the positive real value. Second step is second step is locate the first route a distance of R to the power P by Q found in step number one at an argument of P theta by Q. So theta would be known to you are would be known to you. So you know how to find P theta by Q. So your first complex number that is your first answer the first route I would say would be located at this distance from the origin. From the origin that I forgot to write from the origin and at an argument of P theta by Q. Clear. Then the second route would be located at a distance of same R to the power P by Q from the origin. Okay. From the origin. Let me just do a ditto mark here also from the origin. And the argument would be argument would be the argument of the first route plus two pi by Q. So increase the argument of the previous one by two pi by Q. Are you getting my point and keep doing it. That means if you want to locate your third route. We located at a distance of same R to the power P by Q from the origin. And argument would be P theta by Q now add another two pi by Q to this. That means keep adding keep adding two pi by Q to your previous argument. Keep adding two pi by Q to the argument of the previous route. Okay. But this is how you can easily figure out all the routes and locate them also on the argon plane. So precisely this is what happened over here. So if you see this example, I will demonstrate what I said. My modulus of that complex number was 16. Right. So first I found out 16 to the power one by four. The positive real loot came out to be two. Okay. The argument of this complex number was five. Of course P was one. So P theta by Q will become P was actually one Q was four. So P theta by Q would actually become a pi by four. So if you see first complex number look was located at a modulus of or at a distance of two from the origin and at an angle of pi by four. After that what I did after that I kept on locating the other complex number by just adding two pi by four that is pi by two. So keep adding. Keep adding. Keep adding this to the argument of argument of the previous route. And you'll keep finding out the alpha will get a beta by adding pi by two. So see this is pi by two. Again at a pi by two you get gamma again at a pi by two will get. Sorry. Delta. This is delta my bad. I wrote a gamma here. My mistake. This is delta. Yeah. Again if you add a pi by two will come back to alpha only. So there's no point doing it. Are you getting my point? I will give you another demonstration for it. Don't worry. Huh. Your roots will start repeating. So you don't need need not go any further. Okay. Till you have taken care of your Q roots stop. No need to go any further. Okay. Another observation you will see here is that. Yes. I will slide the screen once again. Give me a minute. If you connect these roots by a. You can say a polygon kind of a structure. You would always realize that you will get a regular polygon. In this case it'll give you a square. So this would be a regular polygon. In this case it is a square. Okay. This case will always be a square. So please note that. No matter whatever is the complex number given to you. And somebody is asking you to locate the roots of. That complex number raised to the power of P by Q. You will always get a Q-sided regular polygon from there. If you connect all your roots on the argument. Always. That's always going to happen. Okay. Is this fine. Okay. They just when you wanted to copy something from. This room. I just want to give you a demonstration also of this, which will make our life very simple. Okay. Please note these things. I mean, these things are not written in books. Okay. So many things I will be telling you from my. An experience mostly. Okay. They will not be mentioned in the books. Done. This diagram. Hmm. This is a diagram you wanted. Done. Okay. Let's take another question. Let's say I asked you this question. Find. Find. Hmm. What should I do? Hmm. Let's say I take a complex number. A root three. Plus I. Okay. And I raise it to the power off. So that'll give me a modulus will be. To the power off. Let's say three by five. Okay. I'm just taking a. Okay. In a hybrid case. So find. Root three plus a race to the power of two by five. Okay. That is your first part of the question. Second part of the question is. Locate. The. Roots on the organ playing. Locate the Roots on the. OK. Now in the interests of time. Okay, so please pay attention. I'll just use the trick to solve this question. And nothing else is required. Ah, yes, madam. P by Q should be in the simplest ratio. I think you are missed out writing this. In the previous slide I had written, maybe here. Look at the top. You see something? Express in the simplest form. Neither if you write one by two as 100 by 200, you will not get 200 answers, right? Okay, let's go back to the question. Yeah. Now the simplest way to get this answer, very easily I will just directly plot it on the diagram. Okay, I will get five answers from here, right? Because this base is a five. Okay, so first find R of the complex number. So Z, let's say it is root three plus I. So what is R of the complex number? R of the complex number is, who will tell me? What is the modulus of Z? Fast, fast, fast, fast. Two, correct. What is the argument of Z? Or theta, what do we call as? Fast, fast, fast, fast. Five by six, correct? So locate your first root. So out of those five roots, the first one will be located at a distance of two to the power three by five, okay? And at an argument of, and at an argument of five by six into three by five. So multiply this with three by five. In fact, this will give you pi by 10, isn't it? So your first complex number will be located at, pi by 10 is almost 18 degrees. So let me vary, you know, small angle here. So 18 degrees somewhat like this, okay? And this distance will be two to the power of three by five. So two to the power three by five, if I calculate it on my calculator, I mean that is not a value that you can remember. So I'll just use my calculator, give me one second. It's around 1.51, 1.51-ish, okay? Anyways, so at an angle of whatever is this value which is pi by 18, sorry, pi by 10, okay? Your first root will be located. So let's say the roots are, I will call those roots by the name of alpha, beta, gamma, delta, and let's say kappa, okay? Kappa another Greek name, okay? Now after this, you don't have to worry anything. You just take two pi divided by the base that is q value, two pi by five, keep increasing this pi by 10 by two pi by five. So two pi by five is almost, if I'm not mistaken, two pi by five is 72 degrees, correct? So keep increasing this 18 degrees by 72 degrees each. So let me write it in terms of degrees only so that you can easily understand it. So now keep increasing by 72 degrees. So 72 degrees will take you to here. And this will be again at the same distance of two to the power three by five. So this will be your position of the second root. So let's say this was alpha, then this will be your beta. Right, so this gap will be 72 degrees gap, clear? And this distance will be two to the power three by five only, correct? Now, when you know the location, you can always write that complex number in the polar form or Euler form, whatever you feel like, okay? Then again from here goes 72 degrees more, okay? So you'll reach here, correct? So again, take a jump of 72 degrees more from here. So you'll reach this position, correct? And this position will be your third root, let's say gamma, correct? Again from there, 72 degrees more, okay? This will be 72 degrees, and you will reach your fourth root, delta. Here, let me write in white, delta, okay? And all of them will be at the same distance, okay? I'm not writing it, but please note that they will all be at a distance of two to the power three by five from the origin, okay? Okay, by the way, one more root is left off that I would have, yeah, let me write it back. Okay, so this would be your third root, let's say kappa, kappa, okay? And this gap will be again a 72 degree gap. If you further increase by 72 degree, you'll reach alpha. This gap will also be 72 degree, are you getting my point? So these roots are positioned uniformly in the space at equidistant, you know, you can say argument difference apart. And they will all be at the same distance of r to the power of p by q from the origin, correct? So looking at this, you can always write alpha, beta, gamma, delta, and kappa in whatever form you feel like. Do you want me to write them down? Okay, let me write them down. Your alpha will be two to the power three by five sys, pi by 10, okay? Beta will be two to the power three by five sys. I just added two by five to this. By the way, two pi by five is four pi by 10. So that makes my addition simpler. So that makes five pi by 10. Five pi by 10 is pi by two, okay? Gamma will be two to the power three by five sys. Add another four pi by 10 to this. By the way, this was five pi by 10. So it'll be nine pi by 10. Now I may exceed the principal argument, okay? So if you want to write it in principle argument, please do so. Delta will be two to the power three by five sys. If I'm not mistaken, this will be 13 pi by 10. And kappa will be two to the power three by five sys, 17 pi by 10, okay? So these will be your five cube roots of root three plus I, sorry, this will be five answers to this particular operation. Is it fine? Right, so directly you can get your result without writing any single word also. Of course, you'll write the relevant things. And as I told you, if you connect them, you will always get a regular polygon. In this case, it will be a regular pentagon. This will be a regular pentagon. Now in JEE and other exams, they may ask you this question like that, okay? They will say, let's say if I were a question center, I will say, if you locate all the values of root three plus I to the power of three by five on a argon thing, okay, you'll get a polygon. Find the side length of the polygon. That would be a question, isn't it? I can ask you to find the side length of the polygon. I can ask you the area of the polygon, right? All these things can be asked. I could also ask you the radius of the circles, circumscribing this polygon, correct? I would ask you the radius of a circle inscribed within this polygon. Anything, any type of questions can be filled, okay? So please be prepared for those kinds of questions also. So I think we have done, you know, concept of the DeMauveret theorem in very detail. Now let us move on to something called the properties of cube roots of unity. So this is something which is important. Normally I do it exactly after my first example, but nevermind. So let's go to the properties of cube roots of unity. Any question here, anybody do let me know, okay? So properties of cube roots of unity. So cube roots of unity you already know are one, okay? One, omega, omega squared. Now since they all have come from cube root of one, that means if you cube them, they will give you a one, isn't it? So of course, one cube is a one. We all know that. And omega cube or let's say omega square cube, et cetera, they will all give you a one. In general, I can say, in general, I can say, if you raise omega to any multiple of three power, of course, n should be the integer here, that will always give you a one. That means omega to the power six, omega to the power nine, 12, et cetera, et cetera, they will all give you a one, okay? If you do omega to the power which is a multiple of three plus one, you'll get a omega back. And if you do omega to the power, a multiple of three plus two, you'll get omega square again, okay? That is obvious, okay? No need to explain anything over there. Tejasini has a question. How to get those answers you just asked? Huh? Which answers? Oh, questions that I just asked. That is up to you, Tejasini. Everything I give you, then what will be your thinking? See, take pleasure in solving the question, not in knowing the answer, right? It's a big statement. Take pleasure in solving the question rather than knowing the answer, right? Means is important than the goal. In learning, means is, you should say, sir, I will try to find it out and I will let you know. That is how it should be, right? Rather than you knowing from me everything. Of course, I try to tell you as much as I can. But these are things which you can always do it yourself, right? It's just basis of trigonometry. So regular polygon, how do you find the side length? You can usually find it by use of trigonometry. Find the area of the polygon also by use of trigonometry. Okay, so try to get those answers on your own. Not a difficult thing. If you are stuck while finding it out, then I'll help you. Next important thing is we already know that omega value is minus half plus i root 3 by 2 and omega square value is minus half minus i root 3 by 2. So looking at it, it's very obvious. So this implies that omega and omega squares are actually conjugates of each other, isn't it? So please note that omega and omega square are conjugates of each other. And since omega cube is one, that means omega into omega square is one, that means omega and omega squares are reciprocal also of each other. Very interesting relation, okay? So not only omega and omega squares are conjugates of each other, they are also reciprocal of each other. So if you do one by omega, you'll get omega square. If you do one by omega square, you'll get an omega, okay? So they are conjugates also and they are reciprocal also of each other. By the way, since you know they are conjugates, let's look into their positioning on the argon plane. So if this is one, this is omega, this is omega square. Okay, please note omega and omega square are conjugates of each other. So they will be mirror-imaged about the real z-axis, correct? And as already discussed with you, if you connect these complex numbers, you will get a regular polygon, okay? In this case, the polygon is a equilateral triangle, okay? So this will be a equilateral triangle. In this case, please understand here, the side length will be root three, okay? So this equilateral triangle, side length will be root three. How did I figure it out? Not a rocket science. I know omega is minus half comma root three by two and omega square is minus half minus root three by two. So the distance between omega and omega square is twice of root three by two, which is root three, okay? So this equilateral triangle, side length is root three. Area would be root three a square by four. I hope everybody knows the area of a equilateral triangle whose side length is given to you, root three a square by four. So in this case, it will be three root three by four, right? Now it is a triangle, I could find it easily. If it is a square, that also you can find easily. If it is a pentagon, you have to construct small, small triangles inside and drop a perpendicular and find it. So I'm just giving you a hint how to do it, okay? Any question with respect to whatever properties I have taken so far? Now a very interesting property which is going to be used several times in many questions. I think yesterday before yesterday I was doing a determinant chapter with your seniors and this property which I'm going to give you now had appeared in one of the questions. Thankfully everybody remembered it. One plus omega plus omega square is a zero. Now this can be achieved from several angles, okay? I think Nikhil was asking a little while ago some of the roots, some of the roots is a zero here. So several ways to basically come to this approach. One is by literally putting the values, your values of omega and omega square. So if I put the values, you will really see that they get cancelled out and leaving you a zero. Leaving you with a zero. So one gets cancelled with a minus half, minus half. I root three by two minus I root three by two gets cancelled, gives you a zero. Another way of looking at it is these equations, okay? Oh sorry, these roots, they come from solving this equation. That is solving z cubed minus one equal to zero, which on factorization gives you something like this, correct? Where the first factor is responsible for one as one of the roots, the second factor is responsible for omega and omega square. In short, omega and omega square appear from solving. So these are, you can say, omega and omega square are roots of z square plus z plus one equal to zero. So if you apply your, you can say quadratic equation formula, this is what you are going to get the values. So here if you see, omega should satisfy this equation, right? So omega square plus omega plus one should be zero because omega is a root of that equation. So this gives you another way of looking at it. Third way of looking at it is something which I already discussed a little while ago. If you see that, this cubic equation doesn't have a z square term. That means if I write it as a proper cubic polynomial equation, where a is your one, b is zero, c is also zero, d is minus one. The sum of the roots, whatever roots I get, that should be minus b by a, which is minus zero by one, which is anyways a zero. So there are three ways of looking at it. So ultimately one plus omega plus omega square will definitely give you a zero. Okay. Another way of looking at it is many people will say, sir, I know this is a geometric series. I hope most of you would have done geometric series in school. So first time is one, common ratio is omega. So I can write it like this as a sum of a geometric series. Now omega cube is a one. So one minus one will be a zero. I mean, so many ways to look at it, the property will stand out irrespective of any path you take. Is it fine? Now this property number three, I will generalize a bit. Sir, you have a habit of generalizing a lot of things. That's how you should actually study. Whenever you solve a question, try to see, can this be generalized? Can I scale this up? Okay. That's how in life we actually scale businesses also. Let's say tomorrow I learn how to make a pot. And I am very impressed with my craftsmanship. Pot is very good. So then I think, can I scale up? Can I make 1000 pots? Can I become rich by that? Okay. That's how businesses also grow. So all that, you can say learning of life. So this JEE is a very simple exam. We are learning for success in life. And your starting point is these problem solvents. So these problem solvents are your small games which you learn to tackle the bigger problems of life. I always keep on saying to my students, JEE, IIT, engineering, medical, the preparation that you are doing is just, you can say, simulation of how do you solve problems later on in your life? Maybe as engineers or as doctors, somebody comes with you, let's say, with something which has not been seen in past. Let's say, Sethu becomes an engineer. And I'm a big builder. I come to him and say, I want this construction to be done. Sethu is saying that nobody has done such a construction in my engineering study so far. So that's what is the challenge for him. That's what the problem for him to solve. That's why he's an engineer, isn't it? Somebody comes to a doctor and says, I have this problem. The doctor may not have seen that disease in his past, but he will basically try to figure out what to do to make this. So as medical students, as engineers, whatever you're trying to pursue, the bigger goal is to solve questions or problems in life. Isn't it? That's how businesses are built. That's how problems in the society are taken care of. Anyways, too much of philosophy. Let's go back to the generalized format. So I will generalize this. I will generalize this. 1 to the power p, omega to the power p, and omega square to the power p, or you can say omega to the power 2p. This will always give you a 0. So if p is not a multiple of 3, whereas this will give you a 3, if p is a multiple of 3. It's a multiple of 3. Now, the second answer is very simple. You'll say, sir, if p is a multiple of 3, then all of them will be like omega raised to the power of multiple of 3. So 1 plus 1 plus 1. 3, simple as that. So how do you get the first one? How do you get the first one? That it is 0 if p is not a multiple of 3. Now, here the trick is very simple. Again, we all have learned our GP. Even if you have not learned your GP, you definitely have seen the formula in the past. OK, without GP also, I can solve it. But let's use a GP if you know it. This is a geometric series. This is a geometric series. I'm sure most of you would have already started sequence and series in your school as well. OK, and the sum is this. Isn't it? We all know this. So treat this as a GP whose first term is 1, common ratio. Common ratio is omega to the power p. Number of terms are 3. So as for this, your sum will be how much? Sum will be a, which is 1. OK, let me write equal to here, as you'll think I'm summing up that. 1, omega to the power of p raised to the power n. So omega to the power np minus 1 by omega to the power p minus 1. Correct? Now, please note, since p is not a multiple of 3, which means omega to the power p cannot be 1, your denominator will be a non-zero term. Whereas your numerator will become 1 minus 1. See, this is going to be, this is going to be, this is going to be, by the way, n is 3 here. Sorry for writing n. Yeah, this is going to be a multiple of 3 anyhow, because 3 into any integer, even though the p integer is not a multiple of 3, let's say p is 2. So 3p will anyways be in multiple of 3 only. So this will anyhow be 1. So this will become 0 divided by a non-zero quantity, which anyways will be a 0, which anyways will be a 0. Correct? Is this fine? Now, many people ask me, sir, can you justify this 3 answer also from the same result? Rather than telling that each one of them will be 1, 1, 1, and hence you'll get a 3. Can I get 3 answer from the same result? Can I use this result? Can I use this result to get 3 answer when p is a multiple of 3? Can I get this answer? Anybody can tell me how to do it? This is here where your thinking comes into picture. Tell, tell, tell, tell, tell. Okay, see if omega, if p is a multiple of 3, omega will be actually 1, right? Okay, let's call it as a variable y. So what do you have here is y cube minus 1 by y minus 1. If you take a limiting case, okay, if you remember I had done a standard algebraic limit which gives you 3, 1 to the power of 3 minus 1. So that gives you 3. So here you have to take a limiting case as y tends to 1. Okay, so this limit is a very interesting concept. In the later part of our chapter, we will try to solve few questions by applying limits. Okay, so this is property number, I think 3 only, 3 generalized form. So let me write 3 here. Any questions, any concerns, do let me know. Fourth property is a simple one. Let's go to the fourth one as well. And since you have learned that the roots of the roots of z square plus z plus 1 equal to 0 are omega and omega square. Can I do this? Can I factorize this term as z minus omega, z minus omega square? Can I do that? Could you wait for two minutes? Yes, yes, why not? Now, fine, done. Anything that you would like to copy? Okay, so guys, please note, I mean, I've used the variable x now. So instead of z, even if you have an x. So up till now, we knew that this is not factorizable. But now after we have done cube roots of unity, we know that this can also be factorized. Okay, that means this expression which you had, which used to stop at this position, now you can actually write it fully as x minus 1, x minus omega, x minus omega square. Okay, you can further generalize this. You can further generalize this and say that any a cube minus b cube expression can be factorized as a minus b, a minus b omega, and a minus b omega square. Please note this down, okay. We also know that minus omega and minus omega square are two of the roots of minus 1 to the power one third, which is as good as saying z cube is minus 1, which means z cube plus 1 is 0, which means if you factorize this, okay, please note, they will actually be roots of this equation, okay, because this will only give a minus 1. So in short, what I wanted to tell you that is minus omega and minus omega square are roots of z square minus z plus 1 equal to 0. That means you could actually factorize this also as z plus omega, z plus omega square. I mean, even if you write it in terms of x, the property doesn't change. So x square plus x, x square minus x plus 1 could always be written as x plus omega, x plus omega square. Now, remember when you used to factorize x cube plus 1, you used to stop at x plus 1 into x square minus x plus 1, but now only to stop there, you can further factorize it if the case requires you to do so, okay. So in general, please note down a cube plus b cube, sorry, I've reached the end of the pace so I cannot drag it any further down. A cube plus b cube can be factorized as a plus b, a plus b omega, a plus b omega square. Is it fine? Any questions, any concerns so far? Now I would like you to work out a question for homework, not now. For homework, prove that a cube plus b cube plus c cube minus 3 a, b, c prove that this could actually be factorized as a plus b plus c into a plus b omega plus c omega square, a plus b omega square c omega, okay. I know most of you already know that this actually can be written like this. A plus b plus c, a square plus b square plus c square minus a, b minus b, c minus c, all you need to show in your homework or try to prove in your homework that this part is obtained by multiplying these two, okay, that's it. That's very easy to prove, you can definitely do it, so please try this out. Many people ask that after proving it should we remember this result? I would say yes, maybe useful somewhere. Can I take a question now on cube roots of unity? Just one question I will take, I will not go into a lot of questions. Clear everyone? Can I go to the question? Okay, let's try this question. If 3 by 2 plus i root 3 by 2 to the power of 50 is 3 to the power 25 x minus i y, find the ordered pair x y or find x and y value, simple as that. Give me a response on the chat box once you're done. Very good, Noel. Okay, let's solve this question. See, let us first try to analyze this expression before we raise it to a power of 50. Is it somehow linked to omega? I have a feeling that if I take a negative root 3 i common, okay from this expression, I will end up getting minus half plus i root 3 by 2. Yes, right? So basically this expression is linked to omega. This is actually your omega, isn't it? Minus half plus i root 3 by 2, that is omega. Okay, that will make my life very easy. So when I raise this to the power of 50, it is as good as raising this to the power of 50. So minus sign will become irrelevant when you're raising it to an even number power. So this will become 3 to the power 25. i to the power 50 if I'm not mistaken is minus 1, isn't it? i to the power 50, 50 is of the nature 4n plus 2. So i to the power 4n plus 2 is normally a minus 1. And omega to the power 50 is omega to the power 48 into omega square. Omega to the power 48 will be anyways be a 1, right? This will be a 1 because 48 is a multiple of 3. So ultimately your answer will be 3 to the power 25 minus 1 into omega square. Minus 1 into omega square will give you half plus i root 3 by 2. So from here if you directly compare, you get your answers pretty straight away. X is half and Y is negative root 3 by 2. Exactly. So I think Noel got it right and Prishnan also got it right. Excellent. Is it okay? All right. So now this cube root of unity concept that we have learned, I'm going to further scale it up, further generalize. What sir? Again generalize. Okay. So let's now talk about nth roots of unity. n could be anything. Okay. Not only 3, it could be any number. 4, 5, 6, 100, 500. Okay. So now let's talk about nth roots of unity. What about the minus sign over there? Where did it, where did that go? It didn't go anywhere. Minus Y is root 3 by 2. So Y is negative root 3 by 2. Yes or no? Sir, compare no? Minus Y is negative root 3 by 2. Sir, Y is negative root 3 by 2. Sorry. Omega square itself is, what is omega square? Omega square itself is minus half minus i root 3 by 2. Minus omega square will become half plus i root 3 by 2. Okay. Okay. Coming back to this concept. Yes. So now we are going to further generalize it and we are now going to talk about nth roots of unity. See the process is very simple. Process is the very same as what we had done to derive the cube roots of unity. So one has a complex number. You can write it as cos 0 plus i sin 0 and I already told you that we can do this operation by D Moivre's theorem. I will not write everything down. I'm just okay using D Moivre's theorem by D Moivre. And K, you start putting as 0, 1 till? Till? Till, till, till, till, till. Who will tell me till? Till? Write on your chat box. Till n minus 1. Right. Challenge me. Correct. Okay. So when you put a 0, you get a 1 only. When you put a 1, you get cos 2 pi by n plus i sin 2 pi by n. Okay. Normally, normally, the books will call this as alpha. Okay. Or let us call this as alpha. Let's not blame books. Let's say I call it as alpha. Then I would like to know, the second, when you put ks2, you get something like this, right? What would you call this as if you're calling this as alpha? Tell me. If you're calling this guy as alpha, then what would you call this guy as? Seethu Sochovir sir. Think again before you answer. See you. If you raise this to some power and that power makes 2 pi by n as 4 pi by n, what should be that power? The power should be 2, right? Yes, absolutely. So it is actually alpha square. Getting it. Okay. All right. So when you put 3, you get 6 pi by n plus i sin 6 pi by n. Again, let me ask you the very same question. If this is alpha, then what will you call this guy? Alpha cube, correct? If you keep on doing this, if you keep on doing this, you would realize that the last guy that I have over here, that means your nth root, that will be alpha to the power n minus 1. Okay. What I want you to do to appreciate here is that they again look like, they again look like a geometric progression, isn't it? 1 alpha, alpha square, alpha cube, alpha to the power 4, alpha to the power n. Okay. So they look like a geometric progression. They look like a geometric progression. Okay. And this is not a surprise because we had already seen a glimpse of it when you were talking about cube roots of unity. 1 omega omega square was basically a specific case out of it. Right. So 1 raised to any 1 by n power, that means any nth root, maybe 5th root, 6th root, 7th root, 100th root, 500th root, whatever. The roots that will come out will always be in a geometric progression. Correct. Where alpha here is, I should write it down here, where alpha is, let's not assume things. Okay. Let's be very specific while we are writing the notes. Let's give me a second. I'll write it properly. Where alpha is, cos 2 pi by n plus i sine 2 pi by n. Okay. Now, many times the questions in the book they will not make it very explicit that it is alpha, alpha squared. They will probably call it as the nth roots of unity is 1 and they will give some number like alpha 1, alpha 2, alpha 3 like that. Now, it is up to you to understand that that alpha 1, alpha 2, alpha 3, etc. that they have given, they're actually in geometric progression. Okay. So many times in the question, they will not make it very obvious. You will have to identify by your prior knowledge. Okay. So please note this down and now we will take up some argon-pane representation. Very simple. So argon-pane representation of n nth roots of unity will be first will be 1. Second root will be again at a distance of 1 and this angle will be this angle will be how much this angle will be 2 pi by n. So this is your alpha. Okay. Next will be located again at a distance of 1 again at a further change of 2 pi by an angle and this will be alpha square. Okay. Another root will be again at n distance of 1 again at a further change of argument by 2 pi by n and so on. So if you keep making it and if you connect them, you will realize that you'll end up getting a regular n-sided polygon which I will not be able to make completely because I cannot sit and draw all the roots. Okay. So this will be a regular regular n-sided polygon. Now here is an opportunity for you to figure out that let there be any n value. How do you find out in terms of n the side length of this regular polygon? Okay. So I have a question for all of you. Find in terms of n in terms of n two things. Side length of the regular polygon regular n-bond second question is find in terms of n area of the regular n-bond. Okay. These two things you figure out for homework. Yes. If n becomes infinity, it'll actually become a circle. Right. Absolutely right. Okay. So just like we took the properties of cube roots of unity, we will also take properties of n-nth roots of unity. Right. So you'll realize that when I take those properties those properties very specific versions are your cube roots of unity. So cube root properties are scaled up to get your properties of n-nth roots of unity. You will soon realize when I start taking them. Can I go to the next slide? Have you all copied this? This expression is very important actually by the way. Whenever you see such kind of an expression, once you should think of, okay, this may be one of the n-nth roots n could be any number. For example, if you see something like this, cos 2 pi by 7 plus i sin 2 pi by 7. Right. If required, or if the question is oriented in such a way that it is related to the seventh root of unity, then this is one of the seventh root of unity. This is one of the seven seventh roots of unity. Okay. So keep this image in your mind because you never know when, you know, any complex number question may require you to apply it. Okay. So just as an example, I, you know, gave it to you. So any expression of the nature cos 2 pi by n plus i sin 2 pi by n is indicative of the fact that that is one of the, that is one of the n-nth roots of unity. Depending upon n whatever is given. For example, n could be a seven in this case. Five, 11, 100, whatever depends on the question. Now we will look into properties of n-n-n-th roots of one or unity. The first property, which is very obvious. Let's say alpha. I just write it down. Let's say let one alpha one, or let's say alpha, alpha square, etc. b n nth roots of unity and if you take alpha the the one which I discussed in the previous slide and you raise it to the power of n or you raise it to the power of any multiple of n that will always give you a one okay p being some integer okay so any multiple of n if you raise it on alpha that will give you a one okay next property next property is very similar to what we had done in our cube root of unity let me not give you like this let me give you like this let one this time I will change the name of okay the roots are the n nth roots of unity so please note in this question while I'm giving you I mentioned alpha one alpha two alpha three place of alpha alpha square alpha cube this is your internal understanding that they are in geometric progression but external examiner when he's asking you a question he need not give you as a geometric progression it is up to you to figure out oh alpha one alpha two alpha three they are in geometric progression so if I call one of them as alpha the other is alpha square alpha cube like that so he will not give you that okay they are in geometric progression it is up to you to figure that out okay so if they are n nth roots of unity please prove that the sum of these n nth roots of unity will always be a zero okay of course one way to prove this is knowing the fact that this is these are actually roots of these are actually roots of z to the power n minus one equal to zero and there is no z to the power n minus one term so coefficient of z to the power n minus one is anyways a zero so that is another way to look at it but if you go by your you can say geometric progression formula so this is actually alpha this is actually alpha square and so on and this is actually alpha to the power n minus one okay so alpha one is actually your alpha then alpha two is alpha square and so on so if you look at your left hand side term it's clear cut a geometric progression okay whose first term is one common ratio is alpha and number of terms are in so if you use your formula it becomes a alpha to the power of n minus one by alpha minus one now remember alpha to the power n from the previous property is a one so this will become one minus one alpha minus one this is a non-zero term because alpha is not a one because one they have already cited separately so this is zero by a non-zero quantity that is always be a zero but this property leads to another you can say you can say sub property which is actually important and i'll be giving you that property there the property is if you sum up cos two k pi by n k from zero to n minus one you will get a zero and if you sum up sine two k pi by n k from zero to n minus one this will also be a zero now you must be wondering how how is this coming and how is it coming from this property okay so first note this down and note these two results down this comes directly from the fact that the sum of the n nth roots of unity is zero so please make a note of this please note down the limits of summation it is going from zero to n minus one in reality it could be some starting from any k value till n consecutive numbers okay zero to n minus one is just one of the cases okay so first note this two result down and then we'll discuss how it comes okay now see look at this result itself this is a that you have written over here this is it all of you pay attention to this result here one is what one is cos zero plus i sine zero right so i will write it in a very fancy way two into zero into pi by n and i sine two into zero into pi by n correct so this one letter you had over here i'm writing it in a very fancy way right at a silo one isn't it the second term alpha one or which is alpha whatever you want to call it is actually cos two pi by n plus i sine two pi by n so that cos two pi by n i will write it as two into one pi by n and i sine two into one pi by n similarly alpha square was cos four pi by n so cos two into two pi by n plus i sine two into two pi by n okay if i continue till your nth term till your last root it is cos two n minus one pi by n plus i sine two into n minus one pi by n okay this is equal to zero and what is zero zero is a complex number which is zero plus i zero isn't it now if you're comparing two complex number remember your comparison of complex numbers that your real part will be equal to the real part and imaginary part will be equal to the imaginary part now what are the real parts in the left hand side who will tell me so it says the real parts are this this this that till this in short the real part is summation i'm just writing it in a notation format so that i don't have to write all of them it is summation two k pi by n from zero to n minus one correct and what are the imaginary part i into same thing summation sine two k pi by n from k equal to zero to n minus one and this is zero plus i zero isn't it so from here it is very evident that summation of cos two k pi by n summed from zero to n minus one that will give us a zero and same will be true for summation of sine two k pi by n from zero to n minus one okay so hence proved simple okay and trust me this result that you see there has been multiple question asked not only in je exam but also in ct as well so this concept is very well known it's not like a high five concept okay don't just be scared by those summation symbols being used it's a very simple concept it just says that if you have angles of the nature two k pi by n and you're summing those sign of those angles or cost of those angles from k equal to zero to n minus one your result will be a zero okay now what question has come on this we will take one question so first note this down and anything that you would like to ask please do so let's take a question can we take a question now okay let's take this question find the value of this summation summation of they've actually kept it as a single term so if you want you can write it like this you can write it as summation sine two k pi or two pi k whatever you want to say 11 from k equal to one till 10 minus i times summation cos two pi k by 11 from k equal to one to 10 uh say to sorry i am reading your message right now which slide you want me to go back to the previous one or the previous to previous one where did you get disconnected never mind i'll share the pdf after the class you can just have a look at it yes please solve this question and give me a response in the chat box okay prisham so i i got an answer from prisham so far how about others okay see here you have been asked to evaluate these two summations correct now we already know sorry we already know that if you sum up sine two k pi by 11 okay from zero to 10 remember n here is 11 so you have to sum from zero to ten n minus one this sum will be zero right but unfortunately the question setter is asking me from one to ten okay prisham note it down i have noted down your correction now the question zero is asking you from one to ten so simple i'll break this summation as first term i will write it uh by putting k as zero manually so that will give me sine zero something is wrong yeah i can write this as sine zero and then summation one to ten so if you're summing from zero to ten zero term i wrote separately and then from one to ten okay so this gives me zero again because this is anyways summation from zero to ten only so sine zero is anyways a zero so basically what i achieve here is what i achieve here is what i achieve here is the fact that this summation is actually a zero so this is a big zero okay now let us try to apply a similar approach to summation of cos two pi k by 11 this is also zero here also if i break it up as first cos zero by by literally putting zero in place of k and summing it up from one to ten okay this should give me a zero the cos zero is actually a one so one for one plus summation from one to ten cos two k pi by 11 is a zero so what does it mean it means summation cos two k pi by 11 from k equal to one till ten is a negative one so this term here is actually a negative one so your answer will be zero minus i negative one answer is of i only this is your answer is it fine any questions any concerns with the solution please do let me know is it fine okay let's move on to the next property i think we had already taken two properties we have taken so let's start with the third one so the third property in fact take it as a question okay take it as a question let one alpha one alpha two till alpha n minus one b n nth roots of unity okay then one to the power p alpha one to the power p alpha two to the power p that till we reach alpha n minus one to the power p okay please prove that this is equal to zero if p is not a multiple of n and it is n if p is a multiple of n okay now would you like to try this out or should i solve it remember we had done a similar property for the cube root of unity as well okay let me solve it so in the interest of time so left hand side if you see these terms that you have you can actually write it as one to the power p alpha to the power p alpha square to the power p alpha cube to the power p because internally you know that they are in geometric progression so knowing the fact that there in gp itself helps you to solve many questions so this is like solving the summation of this question right now you already know it's a geometric progression what is the sum of a geometric progression a alpha to the power p is your common ratio minus one correct so this becomes alpha to the power n p minus one by alpha to the power p minus one okay now remember here if p is not a multiple of three if p is not a multiple of sorry n not three n so we're not talking about n and a truce of unity then what will happen denominator will be non-zero but numerator will be this term will be alpha to the power n p this will be still be a one but alpha to the power p will not be a one so entire sum that is your alpha to the power n p alpha to the power n p minus one by alpha to the power p minus one will become one minus one okay and that will be anyways a zero okay but alpha if alpha is a multiple of n if alpha is a multiple of n then as I told you using limits you can basically answer this so alpha to the power p will be one and let's say alpha to the power p is some y okay so let alpha to the power p be y so this sum that you have alpha to the power n p minus one by alpha to the power p minus one you can evaluate it by writing it as y to the power n minus one by y minus one and you have to take a limiting case of y tending to one because you cannot you cannot substitute y equal to one it will become undefined expression so this is as per our limit result n one to the power n minus one like the way we had done it in our case of cube roots of unity so there you go so please note that if p is a multiple of n you will get n if p is not a multiple of n you'll get a zero anything that you would like to copy from here do so I will show you a very interesting property with respect to this limit thing is it done all right so property number four if one alpha one alpha two dada dada dada till alpha n minus one are n nth roots of unity roots of unity then prove that then prove that in fact prove that is because I want to give this as a question but it's actually a property prove that one minus alpha one one minus alpha two up till one minus alpha n minus one this will give you an n okay now how do I prove this see again I told you in the beginning also that we are trying to find out roots of this equation that means you are trying to find out we are trying to find out or we are trying to say that the roots of this equation are your one alpha one alpha two etc till alpha n minus one so n nth roots of unity are basically the roots of this equation isn't it so if this is the case you would all agree with me that I can actually factorize it like this yes or no if one alpha one alpha two alpha three till alpha n minus one they're all roots of this equation can I say that this polynomial could actually be factorized like this yes or no do you all agree with me or not now see using this result just let's bring z minus one to the denominator okay so what I did I brought the z minus one that is the first factor to the left side now why am I doing all this thing is because if you see the structure of your expression here and the structure of your expression here will actually become the same now does the difference here is here there is a one and here is a z right the only difference is in the required proof we have a one and in this given expression we have a z right so no problem we'll take a limit of we'll put z as one right but please note on the left side you have to take a limiting case because you can't put z as a one okay so here you can put your z as a one that will not make any difference anyways when you evaluate limit for such cases it's a case of substitution only so if you put your z as a one it becomes one minus alpha one one minus alpha two this itself has been asked in several comparative exams and this is n a to the power n minus one which is actually a hence proof so you can see that one vertical of maths which comes from calculus is helping another vertical of mass which is basically coming from algebra okay complex number is an algebra chapter correct so you'll realize that a complex number chapter algebra chapter is taking assistance from a calculus chapter so that is why we say in mathematics and in fact in fact all the science subjects chapters are cross linked so if you don't understand one chapter well you will make mistakes or you'll not be able to utilize in the other chapter is this fine any questions any concerns okay if you have understood this approach let's take a question you need to learn all those properties to be very frank with you these properties will not be asked to you like a property it will be asked to you like a question so think as if you're solving a question rather than looking at a property okay even though I've included it in as a properties of n nth roots of unity but there is high chances that these will become a question itself for unique comparative exams okay setu don't look at it at properties in fact the next question which I'm going to give you it that is also a property but I will give it as a question uh yeah let one alpha one alpha two etc till alpha n minus one are n nth roots of unity and let me write it like this one plus alpha one one plus alpha two till one plus alpha n minus one this expression is uh p when n is even and this expression is q when n is odd okay find find p and q yes any success I think till this step everybody would have done it okay let's bring this z minus one down okay let me write it here itself I don't want to rewrite everything once again let me save my time and energy by putting this z minus one down okay now I want one plus alpha one one plus alpha two so what z value will work over here think about it any suggested z value minus one right absolutely setu so if I put z as minus one what will happen the left hand side will become minus one to the power n minus one by minus two right side will become minus one minus alpha one minus one minus alpha two da da da da till minus one minus alpha n minus one right to be precise the left hand side right hand side expression is minus one to the power n minus one if you take minus minus minus common from all these terms this is what you are going to see correct okay and let's say right side a left side expression I've written now on the right side now here let's try to answer this question uh let me bring this term to the other side minus one so let me divide this term by one by minus one to the power n minus one okay now if your n is even what will happen what will happen to the left hand side left hand side will become now even means this is going to be one minus one zero zero by two zero again and divided by some one or minus one in fact it will be minus one only so that will be zero but if n is even sorry uh when n is odd then what will happen the left hand side will become minus one to the power odd number which is minus two itself so minus two by minus two and this will become minus one to the power even which is actually a one so it's minus two by minus two only and that is actually a one so this number zero is your p and this number one is your q so this is your p and this is your q this itself has come as a question can you believe that so this is actually a property and this has come as a question because the question center also knows that many students will not be remembering these properties so they have a chance to frame this as a question and ask is it fine so we will take a break right now on the other side of the break I will discuss with you a very interesting concept called the kony rotation formula so I think two concepts are left kony rotation formula I think three concepts are left kony rotation formula and we have locus base questions and we have simple complex number equations I think I should be able to finish it by today's class so as of now let's take a break oh now we are going to discuss something important and interesting which is not there in your nct syllabus which is not there in your nct book of course it is there it is required for your j main and j advance exams and this concept is called the kony rotation formula kony rotation formula many places you will find that they would just write it as a rotation theorem okay kony rotation formula or rotation theorem whatever you want to call it a rotation theorem now what is this rotation theorem or what is this kony rotation formula let's try to understand it let's say this is my argon plane okay this is my argon plane and on this argon plane I have three complex numbers let me make them a z1 and let's say we have you know z2 and let's say we have another one z3 okay so I've just taken three complex numbers on this argon plane where if you connect these complex numbers in this way this angle here is theta okay so what has been given you have been given uh three complex numbers z1 z2 z3 and this angle here is mentioned to you as theta now there is an interesting formula called the kony rotation formula which helps you to relate these three complex numbers along with this theta okay so what is that relation let me write that down so that relation is z3 minus z1 by z2 minus z1 is equal to modulus of z3 minus z1 by modulus of z2 minus z1 e to the power i theta okay or you can write the same thing by reciprocating it like z2 minus z1 by z3 minus z1 is equal to modulus of z2 minus z1 by modulus of z3 minus z1 e to the power minus i theta so this formula is what we call as the kony rotation formula okay now first of all we'll derive this formula from where does this formula come right and then we'll see what is the way to remember it okay because it looks to be very complicated because many students ask me year after year so this looks complicated can you tell me a way to remember it so we'll discuss that also and of course we'll also see the application of kony rotation formula now from where does this formula actually come very simple let me make a simple construction over here let me just pull this back and let me put it back in such a way that it doesn't appear to pass through origin okay let's something like this okay and I'll pull this back also okay let's say I call this angle to be theta 1 and I call this angle to be theta 2 and of course this angle is theta itself let me just erase z1 from inside here it is coming in between the diagrams yeah so let's say this is theta okay first of all you would all agree with me when I say z3 minus z1 is a complex number when you subtract two complex numbers you get a complex number and let's say if I want to write it as a Euler form can you tell me if I write it as a Euler form maybe I will put a phi here so that you all can discuss can can you tell me what is going to be r and what is going to be phi so it says that simple r is going to be the distance between z3 z1 which is nothing but which is nothing but modulus of z3 minus z1 correct right and phi is nothing but the angle made by a vector starting from z1 and going towards z3 this vector z1 z3 vector with the positive real z axis and the positive real z axis the angle that it makes is theta 2 so this phi is actually a theta 2 are you all convinced with these two first of all that means in short I can write this as mod z3 minus z1 e to the power i theta 2 let's call this as 1 so are you all convinced with the Euler notation of z3 minus z1 if you have any doubt any concern do let me know okay no issues similarly I can say z2 minus z1 is mod z2 minus z1 e to the power i theta 1 that means modulus is the distance between z1 and z2 that is your r and and your argument will be the angle made by z1 z2 vector with the positive real z axis which is theta 1 as per the diagram how did you get a phi how did I get a phi now if you recall your last class session the argument of a complex number z1 minus z2 let's say I make it on a argon plane so this is z1 and let's say this is z2 or let me make it as z2 here and z1 here so listen to this everybody I'll repeat this again in fact I had discussed this in the the class when I was discussing with you subtraction of complex numbers if you want to find the argument of z z1 minus z2 make a vector starting from z2 to z1 that means make a vector like this okay what angle does this vector make with the real z axis remember for that I asked you a question where I gave you two vectors and I said what is the angle between two vectors and most of you replied it correctly so what is the angle between the vector z2 z1 with the positive real z axis this is the angle right so this angle is what we call as the argument of z1 minus z2 correct Prisham so in the same way if I ask you what is argument of z3 minus z1 you will make a vector from z1 to z3 and see what is the angle that it makes with the positive real z axis which is clearly theta so your phi becomes theta too clear setu is it clear to you as well setu clear right you want one second to digest it so very heavy statements you're throwing today I will take some time to digest me giving you is like you eating but your understanding is like you're digesting okay so having digested this fact let's do one divided by two if you do one divided by two will automatically get z3 so divide the left hand side expressions and divide the right hand side expressions and in right hand side you will get e to the power i theta 2 minus theta 1 correct because when you divide this term by this term as per your exponent loss the angles will get subtracted now what is theta 2 minus theta 1 if you look at from the diagram theta 2 minus theta 1 here will be your theta itself okay so this term that you have over here this term that you have over here this I will replace it with theta only and there you go this is the kony formula which I had you know written on the top of the screen okay okay so this is the proof of the kony formula now the next the other formula which I wrote is just the reciprocal of this result that means if you reciprocate left side and right side terms okay of course reciprocal of e to the power i theta will lead to e to the power minus i theta and that is what is your second result any one of the result if you remember in fact I will tell you what are the difference between the two results when do we use which one okay so this is the proof for this formula and then now I will tell you how to remember it also because the main challenge is sir I always get confused you know whether it will be plus theta or minus theta okay so I'll tell you a way to remember it just a trick if in fact I also follow the same trick to remember this formula see you will be given you will be given three points okay let me name it in the same way as how I named the previous case okay and you would be provided with an angle or you may be knowing an angle let's say this is a kind of a triangle right so you would be knowing one of the angles of the triangle or maybe knowing all the angles of a triangle so let us say let us say the question center has provided me he was kind enough to provide me this angle okay theta now what is the way to remember this kony formula if you want to relate z1 z2 z3 and theta in other words in shorter word if you want to write the kony formula over here you have to do you know one simple activity first this angle is without a direction give it a direction which direction will you like to give it normally the question center is not going to give you any direction he's just going to mention that that angle is theta you need to give a direction to this angle what direction will you like to give it it is your call let me ask somebody say to what direction will you like to give it give some direction to this angle clockwise anticlockwise anticlockwise okay so say to has given an anticlockwise direction to this angle Okay. So when he has given anti-clockwise direction, put a arrow on this particular side just to show it's anti-clockwise. Now this arrow head is hitting a line connecting Z3 and Z1. And this whole scenario is as if like there is a rotation happening with Z1 as the pivot point, isn't it? It is appearing as if like there's a rotation happening, appearing, I'm just talking about appearance. So this is Z1 is like your pivot point. So write the two complex numbers, which you have connected to the arrow head of this angle, writing the pivot complex number as the second complex number like this. So Z3 minus Z1, Z3 and Z1 are the two complex numbers which are hitting this arrow head and Z1 is the pivot point. So write Z3 minus Z1. Pivot should be always be the second one. Okay. Similarly, divide it by the difference of the two complex numbers which are there at the tail of the arrow, which is Z2 and Z1. And again, Z1 will be the pivot point here. So Z1 will be the second complex number. So I'm just telling you a trick to remember the formula. Trick to remember the formula. Now write the modulus of the same two things over here. Okay. And now write E to the power I, if you have taken your angle in a clockwise, anti-clockwise manner, then write a plus theta, which is just a theta. There you go. This basically becomes your formula for the cone, which is matching with whatever you have written over here. Okay. Now, the very same diagram I will draw once again. Let's say somebody says, sir, if I choose clockwise, then what will happen? Okay. Nothing will happen. You'll get the second formula. I'll get the other one of the formula. Okay. So let's say this time for a change, I take clockwise for this theta. So arrow head, first of all, this is your pivot. Whichever is the, whichever are the complex numbers which are connected to the arrow head of this particular angle. Write them first, writing the pivot as the second complex number. So it'll always write Z2 minus Z1. So pivot one should always come after the minus sign. Then the complex number connecting the tail of this angle. So it's Z3 minus Z1. Write down the modulus of both of them. E to the power. Now see, here you have taken in a clockwise manner. This was anti-clockwise. This was anti-clockwise. So if you have taken in a clockwise manner, you have to write minus i theta. This gives you the second formula, which I had given you on the top. See, they are both the same formula. It's just that you're representing it in a different way. Right? So both are same formula. It's just there's another way of writing the same thing. Is it fine? Now you'll remember this Coney formula. Can we see an application of this formula now? So this formula can be used to solve certain questions. OK, so we'll take one of the questions. I will not spend too much time on this. In fact, in the DPP, you can find a lot of questions based on the same. We don't need both of them. We can solve with with any one of them also. OK, let me show you a scenario first. Then you will understand that, OK, do we need to know both of them? Or see, it all depends upon what way you have chosen the what way you have assigned the direction to theta. OK, it is up to you, which which how you want to write the formula. Let me just give you a problem. I mean, instead of talking in general, why not see a problem? OK, problems say things will be clear. OK, let's say this question. Don't worry, I will only solve it. You just have to see the application of the Coney formula. Read this question carefully. This question says complex numbers Z1, Z2, Z3 are vertices of a triangle ABC, respectively, which is a equilateral triangle. You have to show that Z1 square per Z2 square per Z3 square is equal to Z1 Z2 plus Z2 Z3 plus Z3 Z4. By the way, this formula itself is an important property. As a serious JEE aspirant, you should actually remember this also. But again, I'm not asking you to remember things going beyond your you know, remembering capacity. So basically, it's an equilateral triangle and these are the affixes of affixes means vertices. OK, so this is an equilateral triangle. And these all angles are pi by three, pi by three, pi by. OK, now the question setter has given me to prove this. That means if Z1, Z2, Z3 are the vertices of an equilateral triangle, then this relation between Z1, Z2, Z3 will always be true. I have to prove this now. OK, so how will I prove it? For this, I will use my Coney rotation formula. So everybody, please pay attention. Let me drink some water first. OK. Sir, problem is difficult. You are drinking water and all that. No, it's easy. Now, I will call any one of you. To choose a vertex, who will choose a vertex? Noel, can you choose a vertex for me? Any one of the three vertices you can choose. OK, Noel wants to choose vertex A. So Noel, the vertex that you have chosen, I will choose it as my pivot point. So I will assume that the pivot point is Z1. That means this angle pi by three is known to me. OK. Now, let me call somebody else. Who should I call? Karthik Sanoj, Karthik, can you assign a direction to this angle? Clockwise or anticlockwise? It's up to you. Karthik Sanoj, give some direction to this angle. See, here I come to know that the student is not there on the seat. Nidhi Vijay, give some. Sir, clockwise, I couldn't change it also, sir. I'm cortical. No, I didn't receive your message. Sir, it says from me to everyone. Yes, sir, clockwise, put it. OK, OK, Karthik, sorry. So Karthik is there. So clockwise. So Karthik has chosen a clockwise direction like this, right? OK, now, everybody, let us write down the Kony at A. So thinking A is my pivot point. How do I write Kony? So remember the trick which I told you to remember the formula. First, write down the two complex numbers which are hitting the arrowhead of this angle. So that is Z2 and Z1. Z1 being the pivot point, you will write it after the minus sign. OK, then tail of the angle is hitting Z3 and Z1. And Z1 being the pivot point, you will write it like this and write the modulus of the same expression here. Now, since Karthik chose it to be clockwise, you will write minus I pi by 3. Correct. So, Sethu, you had a question. No, which one to use? So it is up to the person who has chosen the direction. So Karthik could have always chosen anti-clockwise. No issues, right? But what I'm doing here is I'm basically giving you a chance to take a call. Whatever call you take, I will accordingly write the expression. OK, now, let me name some other person. Let's name who? Nidhi Vijay. Nidhi, you are there. Z1 is already chosen and we have already written an expression. Now, choose any other pivot point or choose any of the vertex. So Nidhi chooses B, that is Z2 as the pivot point. OK, Nidhi, do one more favor. This angle pi by 3. I hope you can see my cursor rotating on the diagram. Give some direction to that. Anti-clockwise, clockwise. What do you want to choose? Anti-clockwise, OK. So Nidhi chooses anti-clockwise direction for this. Now, depending upon her choice, we will write the Kony formula. So let's do that. So the arrowhead hits Z1 and Z2 and Z2 being the pivot point. This is how I will write it. I hope this expression makes sense. And since her direction of the angle was clockwise, I will write I pi by 3. OK, now, any two of the vertex I have chosen, I have written a Kony formula. Now, from here, I'm going to derive my result. How? Please pay attention. First of all, do you realize that this modulus and this modulus will be equal? Can I cancel this off? You say, sir, how are they equal? See, what does Z2 minus Z1 mod represent? It represents a distance between Z2 and Z1. And what does mod Z3 minus Z1 represent? It represents distance between Z3 and Z1. Aren't they equal? Because this is an equilateral triangle. Similarly, this will also be cancelled off. So can I write this now as and can I write this now as this? Now, I want to strike a relationship between Z1, Z2, Z3. And I don't want this e to the power I pi by 3, etc. to appear. So can I do one thing? Can I multiply one and two? Can I multiply one and two just to get rid of that e to the power i? So when you multiply one and two, the left-hand side terms, let's multiply, they will multiply and these two will become a one. So can I write this as now? Let's take the denominator to the other side. See, numerators here will be Z1 minus Z2 whole square with a minus sign. Denominator, if I multiply, let's see what happens. So this will become minus Z1 square minus Z2 square plus 2 Z1 Z2. This will become Z3 square. This will become minus Z1 Z3 minus Z2 Z3 plus Z1 Z2. One of the Z1 Z2s will get cancelled off or send the square terms to the other side and send the product of 2 to the reverse side. There you go, proved. Okay. Now, how will this question come in competitive exam? They will say if Z1 Z2 are the affixes of an equilateral triangle, then which of the following relation will be correct? And you will find this relation to be sitting in one of your options. Okay. So whenever you realize that there is a situation arising where you have, I can say three complex numbers or any kind, any number of complex numbers and an annual situation is basically getting created there. Okay. Then you can always try to use Kony there. Okay. Or it is recommended that you use Kony there to solve the purpose. Is it fine? Any question, any concerns? Now you can find more problems on in the DPP. Okay. So I will not take too much time because I have to complete this chapter today or today itself. I would love to take more questions, but unfortunately in the interest of time, I am just not taking any more questions. I will now go to the locus or geometry based problems on complex numbers or geometry based problems using complex numbers. Problem using complex numbers. This is a very important concept because in JEE main and JEE advanced exam, in fact, in other competitive exam as well, Manipal and VIT, et cetera, they use, they ask a lot of questions based on this locus or geometry based questions. But there's nothing new to be taught here. Okay. Trust me, there is nothing new to be taught here. Whatever you know related to your geometrical interpretation of modulus, geometrical interpretation of argument, those all are sufficient to crack this kind of a problem. So let me show you some taste of what kind of a problem will be asked based on geometry or locus using complex numbers. Let me give you a scenario. Let's let's directly jump to a question. Okay. Let us say there are two fixed complex numbers. Okay. I'm just giving you the complex numbers themselves. Let's say one of the complex numbers is three plus four, right? Another complex number is minus five plus six, right? Okay. Prove that, prove that a complex number Z, which is equidistant, a complex number Z or prove that the locus of, of Z, which is equidistant, equidistant from three plus four I and minus five plus six I is given by I just write down the locus Z eight plus two I plus Z conjugate eight minus two I plus 36 equal to zero. So locus is not going to leave you. So it was there in coordinate geometry. It was come back again in, in complex numbers. So locus is going to be there in complex number also because, because coordinate geometry and complex numbers go hand in hand because a complex number is like a point, right? So whatever is applicable to coordinate geometry, especially the concept of locus that can also be asked in, in your complex number chapter. Now I will only solve this question. Don't worry about it. You just have to watch. How do I solve this question? Just tell me a simple fact. If I call this as Z one and if I call this as a Z two, how do you write this relation in modulus format? If I want to say the distance of Z from Z one is same as the sense of Z from Z two in modulus language. How do you write it? Please type it on your chat box. If I want to show that distance of our distance between, between Z and Z one is same as the distance between Z and Z two. How do you write this in the language of modulus? That only you tell me. After that I will manage from there. Beautiful. Absolutely. So say to a set. Can I say this is equal to this? Do you all agree with him? Absolutely. And of course, order doesn't matter. You have written Z one minus Z and that to either both are simple. Okay. Now, ideally speaking, this is my locus equation. Okay. Ideally speaking, my question is over here only. But unfortunately, the question setter, which is me, I have given you to prove something which is slightly more complicated. Correct. So what I want to do is basically write the same relation in terms of Z and Z conjugate. So Z conjugate has to come somewhere. So if you want to introduce conjugate, then what will you do? Correct. This is what he corrects. What will you do? You'll say, sir, simple. I will square both sides. Correct. Why are we squaring both sides? Because when you square, I can use the property that mod Z square. Please recall this property. Mod Z square is Z Z conjugate. I can use this here. Correct. So here I can say Z minus Z one times Z minus Z one conjugate will become Z minus Z two Z minus Z two conjugate. Just to bring the conjugate into picture. By the way, conjugate also has a property that you can split the conjugate between the two complex numbers, which we have subtracted here. Okay. Let's multiply. So if you multiply, you'll get Z Z conjugate. By the way, I could have written it as mod Z square also, but let's not, you know, go into that depth. This will be minus Z one conjugate Z minus Z one Z conjugate plus Z one Z one conjugate here also Z Z conjugate minus minus Z two Z conjugate minus Z two conjugate Z plus Z two Z two conjugate. Okay. So let's score off whatever we can score off and let's collect all the Z terms together. Okay. In fact, let's collect it to the other side. Maybe let me take the left hand side to the right. So let's take this term to the other side, this term to the other side, this term to the other side. So it'll become Z Z one conjugate minus Z two conjugate. Then you will have minus Z conjugate Z one minus Z two. And I'll have mod Z two square minus mod Z one square equal to zero. Correct me if I'm wrong. Correct me if I missed out any term. Okay. Now what was my Z one? Sorry. I forgot. What was my Z one Z one was three plus four. I write okay. So let's now use that. So Z one is three plus four. I and that too was minus five plus six. I if I'm not wrong, was it minus five plus six. I minus five plus six. Yes. So let's put it over here. So when I do that Z one conjugate Z one conjugate will be what three minus four I minus Z two conjugate will be minus five minus six. I similarly Z one will be three plus four. I Z two will be minus five plus six. I Z two square Z two. A modular square will be 36 plus 25, which is 61 minus Z one square, which is 25. Let us simplify this. If you simplify this, you get eight plus two. I and this will become eight minus two. I and this will become 36 equal to zero. Is this what we wanted to prove? Is this what you wanted to prove? Yes. This is what we wanted to prove. Right. So this is the question which can be asked to you. See, I have not taught you anything new here. I have just given you the basic use of your geometrical understanding of modulus. Right. Are you getting my point? So this is what we call as the locus-based questions. This is what we call as the locus-based questions. Are you getting my point? Any question in the solution here? Anybody wants to know anything from the solution? Let me know. If it is a multiple choice question, you could have seen both this expression and the initial expression also in your options. So I can make a multiple option correct question also from this. Sir, why don't they give this as an option only? See, again, the question-sitter wants you to test your overall understanding of the topic. Any question? Any concern here? Okay. Now I will give you some drawings to do. Okay. Let's do some drawings here. Since it is locus, locus is a path drawing basically is very important for us. So I'll give you some scenarios in terms of questions. Okay. There is a complex number z1 and there is a complex number z2. These two are fixed complex numbers. Just like your 3 plus 4i and 5 minus 6i, whatever it was in the previous question. So z1, z2 are fixed complex numbers. Okay. There is a complex number z, which is a moving complex number. Okay. It's a moving point, so as to say, but it moves in such a way that it satisfies, z satisfies this locus condition. Okay. Think and answer is a very simple question. So z satisfies the fact that modulus of z minus z1 plus modulus of z minus z2 is equal to modulus of z1 minus z2. Can you draw the locus of z on this diagram? Can you complete this diagram and draw for me? What is the path taken by z so that it always satisfies this locus equation? In fact, I know you cannot draw it on the chat box, but at least you can tell me that this is what I'm going to do to get that idea. I'm waiting for your response. Read it very carefully. The answer is hidden in this locus equation only midpoint. Equipment over. Say it. So, so, so, think don't be a hurry to answer. Line completed. It's not only through midpoint. No, no, no, you have gone off track now. Gali Patriso Tarchukhi. Shalini line between zero and the two awesome Shalini, very, very good. Okay. See, if you take z anywhere in between this, you take here, do you realize that this distance plus this distance is equal to zero. This distance is equal to this distance. Right. And what is this distance? This is mod z minus z one. What is this distance? This is mod z minus z two. And what is this distance? Mod z one minus z two. So isn't it like satisfying this locus condition? Right. So the question that I can ask you this simple version also, you can make a diagram and you'll say that is in the line segment connecting z one, that two between z one, that two, which is the locus equation satisfied by z and one of the options will be this condition on me. Don't worry. I'll give you one more chance at you. Now, same diagram. I'll draw once again. So next question. This is my z one. Okay. This is my z two. Okay. These two are fixed points. These two are fixed points. And now z is a moving point. This satisfies mod z minus z one minus mod z minus z two is equal to mod z one minus z two. So tell me where should z lie or draw the diagram for locus of z here? Tell me, tell me if I'm able to answer this, I will understand your understanding of locus is very good on either side. Can you see beside? I mean left, right, right side. Okay. So say to the same z can lie anywhere here. I mean, if you connect z one and z two and just extend z two, your z can lie anywhere here. Do you agree with him? In fact, I agree with him because here this distance is mod z one, z mod of z minus z one. This distance is mod z minus z two. And the difference of these two distance is the distance between z one minus z two. Excellent. Okay. Now the very same scenario, if I switch it like this, now complete the diagram. If your z satisfies, z is a moving point. Okay. And these two are fixed points. So draw the diagram for this scenario. Tell me this time it will be. Ah, absolutely. So line connecting z one, z two and I know towards the other side. Okay. So please, please note this. They all are basically lying on the line connecting z one, z two. But depending upon which segment of the line they're drawing, this locus equation is changing. So read the question carefully before you end up drawing or before, read the diagram carefully before you end up choosing your locus condition. Okay. This can be a trick point. Okay. Now many people ask me, sir, this locus based question is only having modulus or what argument is not there. Argument is also there. Let me give you a question with argument. Okay. In fact, this time I'll give you a question and you have to draw the diagram. Okay. In fact, last time also we did the same thing. Sorry. Let's say if I ask you, there's a complex number z whose argument is always five by four, always. Okay. So there is a moving point whose argument is always five by four. Can you draw the diagram for z? Can you draw the locus? Draw the locus of z. In fact, when you draw it, just convert it to English language and tell me what you have drawn. In short, just tell me what, what did you do to get the diagram? I've already prepared the argon plane for you, which is prepared time for you to arrive, say to awesome. So say to the same, say to kind of standing clear. Okay. Very good. Say to saying, sir, it will lie on a line, which is making 45 degrees here. Okay. So z can lie anywhere on this line, but mind you, say to and others, it cannot lie on the origin here. That means here you need to put a poll. Sorry, hold the khanapadega kya? Yes, you have to show that whole. Right. Why you have to show that whole because z cannot become origin because origin has undefined argument. You can't say that even that origin is a part of your Lopez diagram, then you are fundamentally incorrect. Then you're saying argument of origin is pi by 4, but the reality is argument of origin is undefined. So here you have to put a whole. Right. So it's a line. Now see line, full line is not required. You can't go down. If you go down, your argument will become minus three pi by four. So only in the first quadrant, they just make a point of this, making 45 degrees and that to origin, you have to carve out by putting that whole. Got the point. Got the point. Okay. Let me let me test you with more questions. Okay. Um, let me give you an inequality question. Argument z is less than pi by four. In fact, okay, less than pi by four. Is it done? See argument of z is less than pi by four. So pi by four line is here. So your complex number will lie in such a way that your argument will be less than pi by four. And please remember your argument will never be, you know, crossing the negative real that access. That means it can lie anywhere below this. Okay. And of course below this axis also, please note that. And you have to draw this as a dotted line, dotted line means it cannot touch that line. Okay. Now many people ask me, sir, why you have done this has dotted because it is less than not less than equal to. So if you take any complex number in this zone, whatever I've shaded, any complex number, you would realize that its argument will always be less than pi by four. Understood. Any questions? Any questions? Okay. Now I will test you on a argument and modulus combined locus questions. Okay. So as a test question, test question, please, please try to solve this question is draw the locus off. Draw the locus off. Z with satisfies. So in the third and for argument will be less than pi. In the third and the fourth quadrant argument, yeah. I think you have to argument negative, sorry, I'm talking to the issue. They say the arguments here are negative, right? So won't it be less than pi by four? Negative angles are less than pi by four. Isn't it? Your argument is between minus pi to pi always, right? Minus pi by pi. You talk about principal argument. No, so here your arguments will be negative. So negative angles are definitely less than pi by four. Okay. Yeah. Okay. So there is a complex number with satisfies. There's a complex number with satisfies this condition. Mod Z is greater than two and argument off argument off. Z is more than equal to pi by two. So can you draw this diagram of a complex number with simultaneously satisfies this condition? I hear my Dukan will be closed. Done. Okay. First of all, mod Z equal to two. How do you draw mod Z equal to two? So this says that the complex number is such that the distance of Z from origin is equal to two. So that will be a circle kind of a thing, isn't it? Like this, having a radius of two, but greater than two means something outside it. Okay. And not including the circle, not including the circle is shown by a dotted circle. You have to make this dotted. Okay. So this is something which is the first locus condition. And I also want the complex number to have an argument greater than pi by two. Greater than pi by two is this zone. So I'll shade it in blue color maybe greater than pi by two will be this one. Greater than pi by two will be this zone. Now your answer will be those region with simultaneously satisfy both these conditions. So which region simultaneously satisfy this condition? Only this part because greater than pi by two is this part. Now I will not go below this line because if I go below this line, it will become a negative argument, which will become lesser than pi by two. So if I have to just draw the actual diagram without showing anything else. Okay. My answer will be everything which is in this part. So your Z can lie anywhere. Of course, I cannot shade all the above. So everything in the third quadrant, second quadrant and outside the circuit. Getting our point. Any questions? Any concerns? Okay. Now I want to talk something about this guy also. Circle equations. Circle equations and complex numbers. So if let's say you have a circle. I'm just making a simple case. If I have a circle whose center is at some complex number and radius is R, then the locus condition that gets satisfied here, everybody would be now comfortable using this. The distance between Z and Z naught will always be R. So this itself is the equation of a circle. Equation of a circle with center at the center at Z naught and radius R. Okay. But you know what? These examiners, they want to complicate this expression. So there's another format of this equation. So what they will do is they will square both the sides. I use, why do they do this? See, they just want to test you from several, you know, perspectives. Okay. So the same equation. Now I'm complicating it. If you expand it, it will become Z, Z conjugate minus Z conjugate Z minus Z naught Z conjugate plus mod Z square minus R squared equal to zero. So this is another version of the very same equation. So this equation and this equations, they both are same things, just written in a complicated manner. So don't be surprised. Don't be surprised if you see this as the equation of a circle given to you. So there was a question, I think in one of the CT papers where they said that find the, so I'll just take an example of that question, find the center radius of the circle, mod Z square minus three minus four I Z minus three plus four I Z conjugate minus minus 25 equal to zero. Please solve this question. Find the center and radius of this circle equation. Simple. Your answer is already there on your screen. You just have to compare these two. That's it. Compare money. Your answer will be in front of you. Done. So it says that simple, I will compare the coefficient of Z conjugate, which is minus Z naught. So here's that conjugate coefficient is this. Correct. So if you compare minus Z naught with minus three plus four, right? It basically gives you Z naught as three plus four, right? So your center will be at three plus four or as a point, you can write it as a three comma four point as a point. Okay. What will be the radius? Very simple. You compare this term with minus 25. So mod Z naught square minus R square is minus 25. Mod Z naught square itself. Since Z naught is this mod Z naught square is itself 25. So that gives you R square as 50. So R is five root two. So this becomes your R. Okay. So such kind of simple questions can be also asked to you under this. Okay. All right. Now a few interesting cases. Please note this down. If a question center gives you, instead of inequality, sorry, instead of equality, he gives you an inequality. I think we have already discussed it. Let's say he gives you greater than, you know, equal to R. Then basically you have to shade the area or you're basically looking at outside this circle area. So everything outside this circle would be a required region. So your Z will lie outside the circle anywhere here. Okay. Whatever. If the question center gives you something like this less than equal to R means you have to consider the area within the circle. I mean, you're all smart enough to figure that out, but still I'm giving this as a note to you so that, you know, you are. So within this, within the circle, the area will be considered. Okay. If the question center gives you a scenario where he mentions that this is between some R1 to R2. Okay. Then basically you have to look at an annular region here. You have to look at an annular region here where the small circle will have a small and big circle. Both will have a center at Z naught. This is going to be R1 and this is going to be R2. Okay. So you are looking at this region. Okay. So your Z is satisfying the condition that it's distance from Z naught is more than R1, but it is less than R2. So it has to be within this region. So all, all these type of questions can be framed. Many times they also frame questions involving. Something like this equal to some theta. Okay. So in such cases, please note, in such cases, please note. One second. Guys, give me a second. Yeah. Sorry. So in such case, what do we do? We first locate the complex number Z naught. Okay. And from Z naught, make a ray in such a way that this angle is theta. Okay. And punch a hole over here because Z naught cannot be included. So you have to put a hole over here. Now, many people say, sir, why is that not will not be included? Because if you put Z as Z naught, it will become argument zero is theta. Again, you're trying to say argument of zero is theta, but argument of zero is undefined. Okay. So such kind of important situations can also be asked. Okay. Okay. Is it fine? Any problem related to this locus-based understanding? So the last concept that we have to talk about is how to solve complex number equations, which is definitely going to be asked in your school exam also. So the last 10 minutes we'll be discussing about, sorry, we'll be discussing about how to solve complex number equations. Can I go to the next slide? Anything that you would like to copy here, ask Claire, do so. One more thing I would like to add here. Many people ask me this question. See, let's say if you get a question like this, draw a complex number. Such that this is, let's say greater than pi by four. Let's say greater than equal to pi by four. How do you draw this? So very simple. One plus I is like your Z naught. Okay. So one plus I is like your one comma one. Correct. Now from here, you have to mark that particular zone where your Z should lie such that if you draw a vector from one comma one to that Z point, its argument should be more than pi by four. Of course, more than pi by four and less than pi. Okay. So what I'm going to do here is I'm going to, I'm going to make an arrow coming from this. Oh, sorry. At an angle of pi by four. Sorry. So from here, I'll make an angle of pi by four. Okay. And I will go max till pi. So I'll make a diagram like this. Okay. So all the complex numbers, all the complex numbers which will lie in this zone. Okay. That will be my answer. Please note that you are not allowed to go below it. If you go below it anyway, let's say you take care, then this complex number will have a negative argument. Please note that the angle that it will make will be negative angle. This will be a negative angle. So can't go below it. Okay. So please ensure you are following these, you know, things while you are drawing the locus. Okay. Last, but not the least part of our chapter is how to solve complex number equations. Actually, this is very easy. You will be given a complex number equation. There you have to follow this simple step. Replace your Z with X plus I Y. Okay. And second step is compare the real parts and imaginary part from that equation in the equation. So once you do this, you'll automatically be able to find your X and Y. The last question you did. Okay. Let me go back to that yesterday. See, if somebody asks you this question, draw the diagram of Z or draw the locus of Z where Z satisfies argument Z minus one plus I is pi by four. Okay. Then make pi by four and make a line parallel to this so that it shows you 90 degrees. So think as if your origin has shifted to one comma one. So now this line, this line is your boundary line. You cannot go below it. Right. And this is your pi by four line. So in short, what I want to say is that this is your pi by four line. And this is now your pipeline. So any complex number that satisfies this condition will always be in the shaded area. It can't go below this line or it can't go towards the right of pi by four. Okay. One more thing. Please note you have to put a whole here as well. Yeah. This point should be a whole. Don't forget it. Why? Because if you are putting that point also in your locus or in your path, means you're saying Z could be one plus I also. That means you're trying to say argument of zero is greater than pi by four, but argument of zero is undefined. There should be a whole over here. Please note this is a whole. Okay. Thank you. Thank you for bringing me back to that slide. I think I had forgotten that. Now I will take a simple demonstration of this, you know, and solve it, solve a question for you. From there, the idea will be clear. Let's say I want to solve this equation. Find all complex numbers with satisfy this condition. Two more Z squared plus Z squared minus five. Okay. So first of all, take your Z as X plus I Y step number one. Okay. Substitute it over here. Mod Z squared will become X squared plus Y squared. Z squared itself will become X squared minus Y squared plus I two X Y minus five. I root three equal to zero instead of zero right zero plus I zero. Okay. Zero plus I zero zero plus zero. All the real parts. Okay. This is your real part together. Take your imaginary parts together. I think imaginary part will only have this and compare it with zero plus I zero. So compare the real with real imaginary with imaginary. So this is your second step compare and imaginary in the equation. So that will automatically give you two equations. One is this and other is this. Now, let's solve these two equations. Let's solve for X and Y in these two equations. So you have to solve for X and Y. So what I'm going to do. Yes, sir. Any question you have said to get a say to please tell me. I Y if you square it, won't you get minus Y squared. You're talking like as if you're doing first class. Oh, sorry, sorry. Yes. Correct. Yeah. Apologies. Okay. Now here. Let's write Y as negative root three by two X and substitute it in this equation. That's the only way I can use to solve it. So let's do that. So I'll get three X squared. Y squared will be three by four X squared. Okay. I hope I am. Yeah. So that will be 12 X to the power four minus 20 X squared plus three equal to zero. I think this is factorizable as minus 18 X squared minus two X squared. Yeah, it's factorizable. So here I will take, take six X squared common. You'll get two X squared minus three. Take minus one common to X squared minus three. So this will be six X squared minus one and two X squared minus three equal to zero. So this implies two things. This is zero and this is zero or or and you can see. So from here X is plus minus one by root six from here X is plus minus root three by root two. Correct. Now, once you've got your X, start finding your Y. So let's put one by root six first. So why will become Y is minus root three by minus root three by two X. So root six will go up. So this will become, if I'm not mistaken, this will give you minus root 18, root 18 is three root two. Correct. Root 18 by two. So this is minus three by root two. So one complex number that you will get is one by root six minus I three by root two. And if I put X as minus one by root six, why will become I think three by root two. So other complex number which I'll get is other complex number which I will get is let me call it a Z one. That two will be minus one by root six plus I three by root two. So that is one complex number as your answer. This is another complex number as your answer or as your you can say the roots of that equation. And the other one will come by using this fact. So if your X is plus root three by root two, Y will become negative root three by two into root three by root two. So that will give you minus one by root two. Correct. So your another complex number that will get formed here is root three by root two minus I one by root two. And if you put X as negative root three by root two, Y will become one by root two. So your another complex number that will be negative root three by negative root two plus I one by root two. So four complex numbers will satisfy this equation. All together four roots will come out. And I think after this step, you should not need my help to solve this. So once you get your X and Y start framing X plus I Y complex numbers from there and whatever complex numbers get formed, they will be your answers. Is this fine? Any questions? So with this, we come to an end of this chapter. Next class, I will be starting with limits. Okay. And I'll just do the limits required for your school stuff. Okay. Let's see how much we are able to complete in the next class. Thank you. Bye bye. Take care.