 Hello students, welcome to cinema gap media channel where we bring for you every day a new problem in the art of problem solving So here we are here. We are again with one problem on complex number as you would have already seen on the thumbnail So let's solve this complex number question and really find out how complex is it? All right So we have three complex numbers z1 z2 z3 says that mod z1 is equal to mod z2 equal to mod z3 Equal to one so all these three complex numbers are Unimodular yeah, all these three complex numbers are unimodular and apart from that We have also been provided with one more information which is right now on your screen, which I am circling with yellow So you can read that information and then we have to find out the Interval in which z1 plus z2 plus z3 modulus lies. All right friends. Let's try to solve this question All right, let's start with the given expression to us z1 square By z2 z3 plus z2 square by z1 z3 plus z3 square by z1 z2 is equal to negative one Okay. Now, our obvious tendency here would be to take the LCM Okay, so let's take the LCM of the denominators, which will clearly pan out to be z1 z2 z3 Giving us z1 cube z2 cube z3 cube Equals to negative one as our given expression. All right. So what we'll do next is I'll take the Denominator to the right-hand side I'll take the denominator to the right-hand side and express the given equation to be like this Now dear friends, our main purpose should be to use as much as possible the expression of z1 plus z2 plus z3 Because that is what has been given to in our question that has been what given to us in our question So wherever possible, I would like to create as many z1 plus z2 plus z3 as possible So in an attempt to do that what I'm going to do. I'm going to I'm going to just Subtract Minus 3 z1 z2 z3 from both the sides. Okay. So first of all this rhymes with our z1 z2 z3 Expression, which I already have on the right side and second of all this can actually be factorized. Yes I can actually factorize z1 cube plus z2 cube plus z3 cube minus 3 z1 z2 z3 as z1 plus z2 plus z3 times z1 square z2 square z3 square minus z1 z2 minus z1 z3 minus z2 z3 and this is equal to Negative times for z1 z2 z3. Okay. Now Let's do even further more simplification here. So we can write we can write our given expression here as Z1 plus z2 plus z3 whole square So what I'm doing here? I'm basically trying to adjust this expression, which I have shown with curly braces as z1 plus z2 plus z3 square minus 2 z1 z2 minus 2 z1 z3 minus 2 z2 z3 Which will finally make the expression look like this minus 3 z1 z2 minus 3 z1 z3 and minus 3 z2 z3 and this is going to be equal to minus 4 z1 z2 z3 In order to write as less as possible I'll make an interim substitution. Let z be z1 plus z2 plus z3 So this is basically an interim substitution Which I am going to do here just to make me write as less as possible Because the more you write the lesser time you get for solving questions in the competitive exams So the entire given expression on the top becomes z times z square minus 3 z1 z2 z2 z3 z1 z3 Okay, so this is equal to minus 4 z1 z2 z3. All right so Let me open the brackets on opening the brackets. We come across this term All right So what I'm going to do next is something which might seem familiar to most of you I'm going to pick up z1 z2 z3 common from these three terms Right, so if I pick up z1 z2 z3 common from these three terms It is very obvious that it'll leave you with 1 by z3 plus 1 by z1 and Plus 1 by z2 now dear students Now as we all know that modulus of z1 is 1 it implies that The square of the modulus of z1 will also be 1 which means in turn that z1 z1 conjugate will also be 1 That implies further that 1 by z1 is z1 conjugate. Can I say similarly? similarly 1 by z2 will be z2 conjugate and 1 by z3 will be z3 conjugate, right? So this gives us a good opportunity to replace this given term in terms of the conjugate of those complex numbers Right, so let us regain back from where we left off and we are now going to write down this as z1 conjugate z2 conjugate z3 conjugate equal to minus 4 z1 z2 z3 Okay Now all of you please pay attention here because this is where some Serious simplifications are going to happen so this expression is z1 plus z2 plus z3 whole conjugate, right and When you have z1 plus z2 plus z3 you can actually replace that with z So I can replace this with a z conjugate So I'll have zz conjugate z1 z2 z3 So I'll repeat this again. So this whole thing this whole thing has been replaced with z conjugate, right? So when I do that I end up getting something of this type So this is equal to minus 4 z1 z2 z3 This gives us z cube minus 3 mod z2 z1 z2 z3 as Minus 4 z1 z2 z3, okay So these students what I'm going to do next is I am going to bring this term to the right side Yes, I'm going to bring this term to the right side. So on doing so I end up getting 3 mod z2 z1 z2 z3 Minus 4 z1 z2 z3, right? So I'm going to take z1 z2 z3 common So that leaves me with 3 mod z2 minus 4 and on the left side We are left with this expression, right? Now what to do next? So may I say that if these two complex symbols are same so would be their moduli So what I do is I equate their modulize, right? So I equate their moduli and when I do that I end up getting something like this So as per the complex number properties of modulus we can write it as Modulus of all these complex numbers individually multiplied to each other, right? So as per the property of complex numbers when you have modulus of z1 into z2 You can safely like write it as a product of modulus z1 into modulus z2 And now all these gentlemen are going to be one each one one one, isn't it? Right, so that leaves us with mod z cube equals to mod 3 of Mod z square minus 4 now This is a very very important expression for us because from here on we are going to do a lot of interesting things So now this mod that you see we can do you know one thing with it. We can take two cases Okay, so one case is let this guy Be already be a positive quantity, okay? So if this guy is already positive, which means you're trying to say mod z square is greater than equal to 4 by 3 Which means you're trying to say mod z is greater than equal to 2 by root 3 Then may I say that the given expression let me call this expression to be e The given expression e reduces to mod z cube equal to 3 mod z square minus 4 Right, so treat this as a cubic equation in mod z treat this as a cubic equation in mod z Right now here if you try to solve this equation one value of mod z that seems to satisfy It is going to be 2. Yes. So mod z equal to 2 Satisfies this given expression, right? It satisfies this given expression. You can check it out 2 cube is 8 minus 3 into 2 square, which is minus 12 plus 4 So 8 minus 12 plus 4 is definitely a zero So mod z equal to 2 is definitely going to meet this criteria and not only that my mod z also Satisfies this criteria that means it is greater than equal to 2 by root 3 So it is an accepted solution my dear friends. All right now What could be the next case the next case is where 3 mod z square minus 4, which is this term could be negative, right? So let's take that as case number 2 So case number 2 could be a case where 3 mod z square minus 4 is negative Which means mod z is less than 2 by root 3 So if that is the case your desired expression e Your desired expression e gets converted to mod z cube equal to 4 minus 3 mod z square because see this is this has been modded, right? And the role of mod is to always give you a positive answer So if 3 z square minus 4 is negative The role of this mod is to return 4 minus 3 z square thereby returning a positive answer to you So right so having said so we can just simplify this expression to be another Cubic polynomial in mod z to be another cubic polynomial in mod z and you can clearly see here is that mod z equal to 1 Simplifies this expression and not only that it also simplifies this restriction that it is less than 2 by root 3 Right, so the possible values of mod z, which I can see is only And by the way, don't forget mod z was our required expression, right? So if you go back to the question if you recall the question, it is basically going to be oh my god You have to scroll up so much. Yeah, it's going to be this expression Which we basically have now figured out towards the end, right? And that expression is going to be taking only two values It's going to take only two values of 1 and 2 so you cannot put this under an interval You have to put a set around this so it's a set containing only two values So guys and girls yes really complex number question can become complicated some at times So that is why this is a potential question for your J main in advanced exams. Thank you so much for watching Be safe and stay healthy