 So when it comes to making notes for studying mathematics, I love to do it in a notebook environment and by that I mean a computational workbook environment. I'm either going to use Julia or Python or the Warframe language and I'm going to use the IDEs for any one of those and I'm going to create a notebook because that allows me to make beautiful type text and also use some code to help me understand or do the problems. Just understand what is going on with the mathematics. So it's really a wonderful for me to have those notebooks really it's really neat to have them as opposed to my terrible handwriting. So as much as I like a thick pencil like a 6b and as much as I like a chalkboard really when it comes to the notes I love them to look nice and so I'm going to create a computer-generated notebook and in that we're going to have some LaTeC code so we have this nice expression of mathematical notation but more importantly we're going to have some code as well because I really don't believe that you should study mathematics any longer without having some code there as well. It really helps an understanding of it. It really helps you to check your work. It just brings a new and a deeper level of understanding as far as the mathematics is concerned. So in this notebook I'm going to just introduce you to complex numbers if you've never seen them before and what we're going to use is a Pluto notebook and we're going to write some Julia code as well. So have a look at this and see if you like it at least you'll get a good introduction to complex numbers. Here we are on a Pluto notebook as you can see everything's already typed out so you don't have to watch me type. So we're talking about complex numbers and you can see here in equation one the natural numbers are a subset of the integers which are a subset of the rational numbers which are a subset of the reals and the real numbers the real number line that is really a subset of the complex numbers. So we're just expanding this set of the real numbers to get to the complex numbers. Now this notebook is within a Julia environment and that Julia environment is in what's described at least in this project.toml file. So I'm saving the address to this toml file as a string and I'm just saving that in a computer variable called file and then we're going to type using pkg. So we're going to use the Julia package manager and then the pkg.activate function and we activate this file or this project.toml file. So all the packages that are installed within this environment will now be available inside of this notebook and we're running that specific environment Julia environment. So if you want to know how to do this how to set up that environment and use it inside of Pluto the link to my video on that is in the description down below. So we're also going to use three other libraries packages and the first one is plots so using plots as well and then plotly is going to be our back end that is the back end I use most often because it produces lovely interactive plots. We also can use Pluto UI just so that I can show you a bit of this user interface Pluto user interface and then one of the inbuilt packages when you install Julia for the first time would be linear algebra and we're going to certainly use that as well. So in case you're not familiar with Pluto notebooks remember all of these things are cells if I hover over the cell you can see that I can create a cell above that by clicking the plus symbol up there or the plus symbol down here it's going to add a symbol and you see the little eyeball there with a line through it which means it's showing or hiding the code at the moment it's hidden so let's show the code and then you can see there so with Pluto remember of course always the code is underneath the execution of that cell is on top so I'm just writing md and then in quotation marks so this denotes the cell as a markdown cell and I'm using a single hash tag symbol there or pound symbol and that means it's going to be rendered as an h1 level tag html tag that is the largest text just as far as normal text is concerned so I'm using that as my title as my title and I can just hide that so it'll make this nice line underneath as well so just for normal text again it's all marked down md and everything goes into a single pair of quotation marks and you'll see how to express LaTeX there you'll see a bit of LaTeX code within dollar symbols that's back slash math bb and then the c within curly braces and that's going to give us this double struct c denoting the set of all complex numbers you see this bold here that's just two underscores before and after some text and that's going to give you bold remember single ones would be italic and then this would all just be LaTeX so again it's just denoting as markdown opening closing quotation marks and then my whole set of the LaTeX code there to have this rendered to the screen and as I would always say that's why I love making my notes inside of some computer language IDE so that we can have these nice lovely neat notes and I just my mind at least responds to that now the imaginary number I mostly has denoted in mathematics although the engineers use the j symbol for that but here we see an equation to the imaginary number which is a horrible horrible name because there's nothing imaginary about it so one of the first things you must do in complex analysis just put out of your head that this is somehow imaginary it's not imaginary at all it's just a poor choice of name but whether they're stuck there's been various other suggestions but this term just just stuck but anyway we see the imaginary number I there and its property it is a number such that if you square it you get negative one that's what it is now we were taught at school that if you square anything it's going to be either zero if the thing that you're squaring is zero or positive everything else even a negative number but that refers to the set of real numbers we're expanding on them now we're now in the world of complex numbers of which real numbers are just a part so indeed there would be numbers that we can square to get a negative value and that number is indeed i if you square it you get negative one in julia to use the this value this element inside of the set of complex numbers the keyword for that is im short for imaginary number if you type im you'll see im is expressed in the code as you see above there so we haven't looked at what a complex number is but the expression here the expression of the code here im squared remember to get a square it's the carrot symbol or shift six on most keyboards so I'm writing im name carrot symbol two that would be i squared and I get this negative one plus zero times the imaginary number so zero times anything even a complex number is still going to be zero so what we left here with actually it's just negative one so i squared is indeed negative one according to the code at least so let's look at this idea of a complex number we see a complex number there in equation three z or z everywhere that's well most places outside of America the US I should say this z equals a plus bi and a common alternative which we will probably use most often in this notebook is x plus i y now you just see bi and then i y it's just the other way around there's there's a commutative property there so it doesn't matter which way around you write it but it's common if you use x and y's that you write i y and if you use anything else it's bi anything else i so bi now what is this a and b and x and y they're just two x and y they're just two real numbers they're both elements of the real numbers the x or the a that is called the real part of this imaginary of this complex number z and y or b those are the imaginary parts so the real part x and a and the imaginary part let's be a y and we usually denote that as you can see r e of z and and the i m of z r e of z i m of z so that is a complex number and some text you'll also see this notation just wanted to show you that z equals an a comma b that's still the real and the imaginary parts of this complex number now i promise i'm going to show you just a little bit of why one of the reasons why i think people like pluto is is the pluto ui they can do this with without the macro code but since we've imported that package pluto ui we can just use the macro at bind and then we're going to have a variable called a and what we want that variable to be is the value of a slider object so slider and i'm using unit range a negative five to five and because we're not using a step size there the default step size will be one so it'll be negative five negative four negative three all the way to three four five and that gets expressed as a slider there and we see we do the same thing for b and i can drag the slider and it's going to give me different values because let me move it up see what's happening down here so think about a pluto notebook everything is live so wherever i change some code if it's the variable is involved in another cell that's going to update at the same time so i've created this imaginary number here or this complex number z one and i said it's a plus bi and that a is a computer variable and we've created it up there so at the moment both a and b according to the sliders are negative five so our complex number at the moment is negative five minus five i and as i drag say the a you see that updates immediately that updates immediately in three and four and five there we go now it's five minus five i m or five plus four i m so it's as simple as that and you can well imagine i suppose a couple of uses of just these sliders and they're more than just the slider objects and pluto ui so give it a give it a look now there are three functions inside of julia that are useful when it comes to these complex numbers the first one is real so if i pass in a complex number as an argument to the real functions it's going to give me the real part back and again as i drag the slider you'll see my number updates it's now one plus four i m so the real part is one and then i m a g for imaginary pass the complex number as an argument and we can see that or you can return it as a tuple using the r e i m function r e i m function and it's going to give you a tuple back of the real and imaginary parts so that's very simple now one thing we wanted to do in this section is just a bit of simple simple arithmetic so let's create two complex numbers we're going to have z one and z two and if we see any equation three so they've each got their own real part and their own imaginary part if i add them well we just you know we've added the z one z plus z sub one z sub two on the left hand side so we're just going to add these two things on the right hand side so that'll be x sub one plus i y sub one plus x sub two plus i y sub two and all we're going to do now is we're just going to group the real and imaginary parts so we see if we add two complex numbers we're just going to add the real parts and we add the imaginary parts and the imaginary parts just get multiplied by i as simple as that so let's just have a look at this we're going to create a second complex number z two remember z one is still going to be bound to those two sliders at the moment for me it's one plus four i am we're going to create z two that's three minus three i am and then i'm going to say z one plus z two so at the moment my real parts are one and three three plus one's four and four and minus three that's one so indeed the solution would be four plus one i so very very simple indeed more of more interest let's look at multiplication and we're going to do it the same as we did with algebra if we have x a plus b times m m plus n whatever we're just going to multiply it up by those rules that you learned at school so here we have let's look at the top here z one times z two so that's x sub one plus y i y sub one times x sub two plus i y sub two so x sub one times the x sub two goes in the second line then x sub one times i y sub two so there'll be i x sub one y sub two plus then the i y sub one times the x sub two just rearranging things so it looks neat so there'll be i x sub two y sub one and in the end we'll have a plus i squared y one y two and all we're going to do now we're going to group the real and imaginary parts and we also can remember that this very last expression i squared y one y two is actually a real number because i squared is minus one so we're going to have x sub one y sub x one x sub two minus y sub one y sub two plus then i multiplied by the expression as you can see this so very very easy no problems whatsoever let's just look at an example i have z sub three and z sub four there z sub three being two plus three i and z sub four being negative two minus i so if i multiply these out the real part is going to be two times negative two that's negative four and then at the end it's going to be three times negative one that's negative three and i squared which is negative one so that gives us a positive three so it's minus four positive three which leaves us with a negative one and then the i's are going to be a two i and a negative two i and a negative six i that gives us negative eight i so there we created code and if we multiply that out indeed we get negative one minus eight i no problems whatsoever so write yourself out some and just test test it by writing uh writing some code now we're supposed to get to division but of course division is not going to work for us we'll have to define um or make a created definition for for division of complex numbers and to do that we first have to introduce this notion of a complex conjugate so if a complex conjugate as you can see here z bar a bar over the top sometimes there's a superscript asterisk but let's use the common notation which is just the bar over the z of a complex number z that's defined in equation six so z equals x plus i y then z bar is x minus i y so all we're going to do we're going to flip the sign on the imaginary part or at least multiply it by the real number negative one so remember what z four was z four was negative two minus i and if i take the conjugate of that complex conjugate of which the function is just c o and j of z four i get negative two plus one i because negative one if we flip the sign or multiply by negative one becomes a positive so it's the complex conjugate so for interest sake let's just see what happens if we multiply a complex number by complex conjugate and you can see there very quickly that we're going to end up because these i's on there and they're going to cancel out minus i x y plus i x y they cancel out i squared is a negative one and that becomes times this negative is a positive so we get x squared plus y squared so we have the fact that a complex number times its complex conjugate gives us a real number so there's a little example for you we take z four and we're just going to multiply it by its complex conjugate z four so that leaves us with five and you can check on the result there that would be z four times the conjugate of z four and that gives us five because it's a real number because we have zero i so now we can find a fine division and this is what we what we're going to do if we have two complex numbers z one and z sub two such that z sub two is not zero it's not zero plus zero i we can then we can do division and what we do is we multiply by the complex conjugate of the denominator over the complex conjugate of the denominator and still in complex analysis dividing a complex number by itself that just gives you one anything divided by itself is one except of course for zero dividing by zero is not defined so if i have x plus x sub one plus i y sub one that's my z sub one divided by x sub two plus i y sub two i'm going to multiply just by the denominator's complex conjugate over itself and that's what you're going to get right the bottom expression here in equation nine that the at least the denominator is a real number and then the numerator is still going to be a complex number and now as you can see in the last expression last equation here in or last line in equation one the denominator is a real number and the numerator is still going to be a complex number so let's have this little example of z five and z six and you can see three minus three i is z sub six so we're going to multiply by three plus three i divided by three plus three i and then the denominator comes the real number and the numerator is a complex number and we can again just check that out with code creating z five creating z six and saying z five divided by z six and gives us this negative zero point one six six all the way plus one plus six six all the way i and just to show you if i take the solution here negative three plus 21 i divided by 18 i get exactly the same result so that is indeed the numerical representation or an approximation of the of the fraction there that we see as far as this is concerned you could also write it negative three over 18 plus 21 over 18 i so that brings us neatly into this idea of you know we have this real number line so we how do we visualize this extension to the complex numbers seeing that the real numbers is a subset of the complex numbers well the real line we still have and then this diagram which is called the argent diagram we see an orange that's the real axis with all the real numbers but perpendicular to that we create another axis and we call that the imaginary axis just as we had an x and a y axis but in Cartesian coordinates remember both the x-axis and the y-axis they independent they're just independent real real number lines yeah we and if we have any if we have any point on the Cartesian plane that refers to two real numbers and usually remember I don't parentheses and it's one number the number the x comma y value of that point but this is not what we have here we have the real part and the imaginary part so if we have a complex number now we can plot it on this argent diagram based on its real value and it's imagine it's real part it's imaginary part because both of those are real numbers and that's exactly what happens here so we can see the value one comma one here so on the real number that we won on the imaginary axis that we won so that's our that's our complex number there one plus i and here's another complex number that's two negative one so it's two minus i but you can also see something else we can also view these things of course as vectors if we have the origin of the vector or here at the origin and then the vector extends to the complex number because if you think about that this allows us you know a rich understanding of complex numbers through a bit of geometry and we have this idea of right triangles because if this is the number one comma one that means this length on the x axis is one the length on this imaginary axis is one we have a right angle triangle so we can have this notion of the length of this complex number or the length of this vector and we're doing the talk about the length of a complex number it's called the modulus and we'll get to that and there's also this angle that the vector makes with the positive x axis or the positive real axis we should say so there's a lovely geometry and trigonometry involved here with these complex numbers if we view them as these points on the argn diagram or the complex plane in any way there's the bit of code and as i said that's why i like the plotly back end for plots because we can now we have this interactive idea of this interaction at least and we can zoom in and now we really zoomed in and then we can just go back home etc and we'll just save this image as a png file so let's look at this idea of the magnitude of a complex number but you've seen here now it's not that's going to be nothing other than than just Pythagorean theorem basically because that complex number will just be the hypotenuse of a right angle triangle nothing nothing other than that so this magnitude of a complex number we call it a modulus as you can see here we define it in equation 10 and we we write it with the straight lines around the complex number and that would be positive the square root of x squared plus y squared and remember how we got x squared plus y squared yeah that was just the complex number divided by multiplied by its complex conjugate and we take the square root of that and what you can see the x squared plus y squared square root of that again that just half-spect the Pythagorean theorem doesn't it so it all binds lovely nicely together and of course we're only interested in the positive part because we're interested in the length of this thing so if we have z sub five remember that was three plus four i so that's going to be three squared plus four squared that's 25 and the square root is 25 is five and that's what we would expect from a right triangle with a base of three and a height of four no problem there and there we do it we say the square root of the real part of z five squared plus the imaginary part of z five squared we take the square root of all of that we get five or we take the square root of z multiplied by its complex conjugate same story and then there's also the abs function for absolute and that's going to give us the modulus of repass as argument the complex number so that'd be three ways for you there to calculate the modulus of the complex number so now that we know what the length is we can suppose also look at that angle that it makes for the comp with the positive real axis and we call that angle theta most of the time and remember if we go around and this two pi radians or 360 degrees from the positive real axis counterclockwise all the way around there's nothing stopping us there's no barrier there we can just go round and around and around and every time we add 360 degrees or two pi radians so that first time we go before we go round twice that angle that first time round anyway would be what we call the principal argument and if you study a bit of complex analysis you know we'll go beyond the principal argument and through just right angle triangle trigonometry it's very easy to see that this angle is going to be the arc tangent of the complex part divided by the real part it's just a simple right angle triangle taking the tangent of that angle tangent remembers opposite divided by adjacent you already know the opposite and adjacent side because that's the real and imaginary parts so opposite being y that's the imaginary part divided by x and that is going to be the real part and they just went to sake i'll show you the arc tangent function there's the code to do the arc tangent function and so there we go so let's take this z sub seven it's one plus square root of three and i want for this example i want this to be expressed this angle to be expressed in degrees so in radians at least that's going to be the arc tangent of square root of three over one the imaginary part divided by the real part and the arc tangent of square root of three well that's just going to leave us as fast degrees as concerned at a 60 degree angle remember that's counterclockwise we're dealing with the first quadrant here with the number one plus one is positive square root of three is positive so that's going to be in the first quadrant of the arc and diagram of complex plane and that's going to be 60 degrees so if we do this z one as one plus the square root of three times i and we take the arc tangent of the imaginary part divided by the real part which you can see the a tan is the arc tangent function or the inverse tangent function and i wrap that all in the red two degree function which is going to take radians as input and output degrees and bar the the round off error we see that indeed the solution is 60 degrees i can also wrap all of that in the round function and that's just going to round it to that integer which is which is 60 degrees and imagine then we just have what i what we should go through at least is just just considering all four of the quadrants in the arc and diagram so we're going to keep things very simple we start with our first one z equals one plus i so it's one down the real axis one up the imaginary axis makes a nice pi over four principal argument going counterclockwise from the positive imagine a positive real axis and if we do that we can use the angle function so instead of here we had the arc tangent of the imaginary part divided by the real part there's also i should have said that the angle function that's going to do exactly the same thing for us so we're going to take the angle of one plus i m and we wrap that in radians to degrees and we see we get the solution of up here 45 just as we would expect now if we go to the second quadrant so that is going to be negative on the real axis but still positive on the imaginary axis so minus one plus i you can really visualize in your head where this is going to be otherwise just create a little graph as we did up there just copy and paste that and change it a bit and that gives us as we would expect 135 degrees because it's counting the angles from the positive real axis counterclockwise till we get to this complex number vector representation and if we then go to the third quadrant so there'll be both negative on the real part and the imaginary parts a minus one minus i we see we get an angle of negative 135 and this is by design so in the third and fourth quadrants we're going to go from the positive real axis clockwise so going down and then to the left and that's how we get to 135 but it's now minus 135 and then the fourth quadrant which will be something like one minus i m or any representative number in that complex number in that quadrant i'm just choosing easy ones we see an angle of negative 45 degrees so in those two quadrants it's going to go clockwise downwards from the positive real axis and get these negative values but as i said we can keep going so a number like one plus i although it has a principal argument of pi over four radians of 45 degrees we can add go around another two pi radians or 360 degrees so we'll be pi over four plus another two pi that's nine over four pi if you add those two and if we pass that to the tangent function and round it we see we still get this value of one so if we take the arc tangent of one radians two degrees we get 45 degrees we get 45 degrees so even though we've gone around more than once it still ends up we're gonna end up with the same angle the same angle it's going to repeat then repeat and repeat if you think about the way that the tangent function works so that's the geometry and trigonometry just an introduction to that as far as complex numbers concerned it's really intuitive so let's get to Euler's equation and that's very nice because now we're going to use to get still more trigonometry and we're going to gain an even deeper insight into this idea of a complex number and our again diagram because if you look at the cosine let's just imagine the first quadrant so there's in this nice little vector that goes out into the first quadrant so a complex number in the first quadrant and you take the cosine of that angle that it makes for the positive real positive real axis and the cosine it's just adjacent divided by hypotenuse and if we make this hypotenuse r or the length the modulus of our complex number if then we make that r so the cosine of the angle is just going to be x that is the real part the length of the real part divided by the hypotenuse that means we get another expression for this real part it is the modulus times the cosine of the principal argument that's a mouthful but it's easy to see there and the sine of that angle of that complex number is just going to be the modulus times the sine of that angle that would be y y is going to be or at least the imaginary part you should be clear just look at that equation 12 so the imaginary part is the modulus times the sine of the argument and the real part is the modulus times the cosine of that angle the principal angle now let's just check we just said well the Pythagorean theorem works for us but we can also just sort of derive it here if we use the trigonometric identity that the cosine square of an angle plus the sine square of an angle or the sine square of that angle plus the cosine square of that angle that's just one because if I now use this new way of writing the real and imaginary parts being r cosine theta and r sine theta if I have x squared plus y squared that is going to be r squared cosine squared of theta plus r squared sine squared of theta I take the r squared out as a common factor and I have this cosine squared theta plus sine square theta which according to the most well known I suppose when you learn first the trigonometric identity you learn about first that's just one so x squared plus y squared is r squared or r equals positive the square root of x squared plus y squared so it's a no-brainer there so let's take this equation 14 and I'm rewriting x plus i y my x is now r cosine theta my y is our sine theta and I have this i I see that it can take out r as a common factor so I have this brand new way of writing any complex number it's the modulus times the sum of the cosine of the principal argument plus i times the sine of the principal argument that's great so from there we're going to make this little big jump here in equation 15 because what we're going to start off with is just a series expansion this approximation of the cosine of an angle and the sine of an angle and you'll remember that from first your calculus this series expansion of the cosine of an angle and the sine of an angle and you see them written out there except for the sine of angle on the left hand side with multiplied by i which means every little expression component on the right hand side has to be multiplied by i as well and then you also see we take the exponent theta or Euler's number to the power of theta and you see the series expansion there and now instead of just theta I'm going to say e to the power i theta so everywhere where I see theta I replace that with i theta so all I'm doing on the left hand side so we're doing it on the right hand side replacing theta with i theta and now it's very neat because all these numerators have i squared in them then i cubed i to the power of four and of course we can work all that out i squared is negative one so we see this e to the power i theta that becomes one plus i theta and then minus theta squared over two factorial so those exclamation marks is factorial and so I have this series expansion of e to the power i theta and what I notice that you know every second one is just a real number and every other second one contains a complex number so let me just take out this uh separate these two out so that I only have these real number expansion and then the expansion that has i in it and what you'd notice there in these two final sections is that I have the series expansion of cosine theta there and I have the series expansion of i sine theta on the left side so I can just substitute that back in so e to the i pi is cosine of theta plus i sine of theta that's wonderful and if cosine of theta plus i sine of theta equals e to the i theta I can replace that in this z equals r cosine of theta plus i sine of theta equation of hours that means I get this equation 16 Euler's equation z equals r times e to the power i theta so I can write a complex number as the modulus times Euler's number to the power i times the principal argument that is wonderful and now it's just to be very bold let's just pick a complex number that's on the unit circle so the modulus is one and we're going to on that on that circle we're going to choose pi radians now think a little bit in your head or draw down paper if I have the complex plane and I have a principal argument of theta radians that flops me counterclockwise all the way over to the negative real axis and I have a modulus of one well that's just going to be the complex number negative one plus zero i or the real number minus one that's all I have there and I have this fact then that e to the i pi if I then write it out instead of thinking about the just the image in our head of this argn diagram if we write this out arg is now one so that disappears cosine of pi plus i times the sine of pi well we know from before that that is negative one so e to the i pi equals negative one or e to the i pi plus one equals zero and that's just phenomenal this idea this connection this deep connection between the oilers number e and pi this is beautiful connection with the two such as if you write it in this configuration e to the power i theta I should say e to the power i pi this is negative one lovely stuff so this brings us then with a new way that we can multiply and divide two complex numbers you see equation 18 and 19 so equation 18 we multiplying z sub one and z sub two and we're writing it using this oilers using an oilers form there r sub one e to the power i theta sub one times r sub two e to the power i theta sub two so we see that's all just four terms multiplied by each other so we can rearrange a little bit the commutative property so r sub one times r sub two times e and then with the law of exponents remember if we multiply these two we can just add them and I've added them there in the second line and I've taken i out as a common factor so with multiplication we multiply the two moduli and we add the two principal arguments and we raise multiply that by i and use that in the in exponentiation and then of course it becomes very easy if we have if we multiply more than two complex numbers we'll just keep on multiplying the moduli and we'll keep on adding the principal arguments no problem there and we see the division there as long as z two is not all zeros not zero plus zero i the real and imaginary parts can't be zero we have this idea of dividing the two moduli and we subtract the second from the first as far as the principle argument is concerned which brings us by this lovely theorem and I went to try and butcher that name look it up on youtube or it's actually quite funny to see how many pronunciations they are but if you go to google and just try and look for the official pronunciation it'll do a much better job than me and the theorem is written there in equation 20 it says if I take a complex number to any power that is going to be the same as taking the real the modulus to that power and this idea of the angle angle form so cosine of theta plus i sine of theta and I and I take that to the power n and what we're suggesting that's the very same thing then the modulus to the power n times cosine of n times theta plus i times n sine theta but somehow as if by magic I can bring that the power into the angles as well now the proof of that is be simple we're going to prove it through and prove it by mathematical induction simple induction so we're going to use the fact that the cosine of the addition of two angles can be written as the cosine of the first angle times the cosine of the second angle minus the sine of the first angle times the sine of the second angle and we also have this idea of if we add two angles and we take the sine of that that's going to be the cosine of the first angle sine of the second angle plus sine of the first angle cosine of the second angle no problem so let's look at the base case n equals one so if I just put n equals one there we have our cosine theta plus i sine of theta as far as z is concerned so we a okay there for at least for one works if we if we say z to the power one and everything is to the power one and it's one times theta and for cosine and one times theta for the sine I mean it just all works out for a base case one now we do this induction hypothesis we say we assume that the theorem is true for n equals k and show that it's true for k plus or n equals k plus one so I have things there z to the power k we don't know what k is that's going to be r to the power k as we mentioned and then the cosine of k theta plus the sine of k theta times our cosine theta plus i sine of theta because if we have z to the power k there and we say z to the power k plus one that's all we do is we have z to the power k times z because if we have it you can see it right here z to the power k it's not going to highlight for me but z to the power k times z well we just add the the exponent so that we k plus one and on the right hand side you can see what it is let's do the power r r to the power k we have this notion of multiplication of the angle as far as cosine and sine is concerned and all of that we're just going to multiply by the modulus times r sine our cosine of theta plus i sine of theta and you see there how we we expand that and eventually we get to this idea of r to the power k plus one equals indeed r to the power z to the power k plus one equals r to the power k plus one times the cosine of k plus one theta plus i times k plus one theta so if we assume that z to the power k is true we see that z to the power k plus one works out for us as well and all the dominoes fall because if we assume one is true and we show that the next one is true if we then start with the base case which we're shown to be true that means the next one has to be true and if that one's true the next one has to be and all the dominoes fall and that's mathematical induction so let's have a look at this idea then of taking the fourth roots of a number one well now you didn't know there didn't know before at least that there are actually four roots there's four numbers that you can multiply by itself four times that is going to give you one so the fourth roots of one or we can just write it as one to the power quarter which means n is a quarter it also means that r equals one theta is zero this is just one so it's this dot on the real real part of the complex plane the real axis and then we have this fact that remember that if you take one and you raise it to any power except of course zero well one to the power zero is actually also just one isn't it and so that's just going to be one so we can simplify that first line a little bit as well because if we write it out now it's going to be the cosine of that angle plus y sine of that angle and because one to the power quarter is just one that becomes a one in line two there and I have this idea of theta plus two k theta divided by four and theta plus two k pi divided by four and you can see there the sequence of values for k this means you know the fact that the theta equals zero so we can leave that out we are left with this k pi over two and k pi over two as far as cosine and sine is concerned and if you start with k equals zero that's going to be cosine of pi over two plus sine of pi over two and that just leaves us with one and if k equals one two I should say or it start with zero then one then two then three you see the four numbers that you get there one i negative one negative i and if you take any of those values and you raise it to the power four you're going to get the value one check it out for yourself now the last part of this notebook is just revisiting the sine and cosine functions and what I do here in this last section that you can read on your own that's very neatly typed out for you there is that we can rewrite complex numbers or the cosine at least of an angle we can rewrite it as you can see in equation two cosine of an angle is e to the power i that angle plus e to the power negative i times that angle divided by two and then the sine of an angle we can write like that and we just divide it by two i remember to get rid of this complex part in the denominator we can multiply by negative two i over negative two i okay which brings us here by just Euler's notation z to the power n remember that z to the power n it's cosine in theta plus i n theta and if it's z to the power negative n remember the cosine function here if you take the negative of a negative angle it's still just going to be the cosine of a positive angle and then minus i times sine of n times theta and if we add them there and we remember from equation 25 here when n equals one we get two cosine theta from equation 23 and in the second form two i sine of theta and we can use this fact to find some other expressions say for instance in example seven cosine to the power four of theta and we would just just this fact that we've just looked at now these equations up here 23 24 25 we just use them so that we get this idea of cosine to the power four can be written as one of eight times cosine of four times that angle and it goes the other way around as well when we want to write something as the power of these angles it's very easy to do as well and you can follow the steps here in example eight and look at the code to verify that for yourself the last thing i want you to read here as well is about the dot and the cross product and how to calculate them and what they mean to us and you can see all the code there and then how to work out how to do this what you are going to need as well is this idea of the determinant of taking this matrix and as far as the cross product is concerned so you can certainly read through that and that's it that's your introduction to complex numbers and we make very neat notes using a pluto notebook and we make life very easy for ourselves look at yeah we write this cross and we take the cross product of these two complex numbers three plus four i and two plus four i and lo and behold we get this solution right there zero zero four and you can read why that is so why the cross product do we have the solution that is perpendicular to the argon diagram that just brings us with two very important things to note when it comes to the dot and the cross product if you take the dot product of two vectors two complex numbers and they are zero the dot product zero then it means either it just really means that the two vectors that make other complex numbers are perpendicular to each other and if we take the magnitude of the cross product if that is zero it means these two values are perpendicular to each other so read this last part that's quite fascinating the dot and the cross product of complex numbers so learn how to write this uh bit of code how to do a bit of lartech just write normal english sentences in some of these code cells give it a go your work will be so much neater you'll understand it better it looks good these solutions are very well written in as far as it's very neat to do it in to do your mathematics homework or at least your notes inside of a notebook environment such as this i hope you enjoyed this video give it a thumbs up subscribe to this channel that is the most important thing i bring out a variety of educational resources it's just something that i love to do when i'm not in the hospital doing my day and sometimes night job as a surgeon i love to make these please appreciate it by by joining this this channel and this community and leave leave a comment down below as well so i've got some videos on julia i've even got a course on course sierra that you can get official certificate from the university of cape town if you want to learn some julia so have a look on course sierra.org the biggest massive open online course platform in the world and look up julia for scientific computing and you'll see the course there by myself and one of my colleagues in the mathematics department on velori and that's a fantastic course as well otherwise check out all my videos here on youtube and learn about using julia and do your mathematics at least the numerical part with julia