Complex Algebra Fundamentals | Dmitri N. | Skillshare
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16 Lessons (1h 5m)
    • 1. Overview

      1:45
    • 2. Real and Imaginary Parts

      2:17
    • 3. Addition and Multiplication

      5:24
    • 4. Basic Algebraic Properties

      5:30
    • 5. Vector Representation

      4:48
    • 6. Modulus

      4:06
    • 7. Complex Conjugate

      4:50
    • 8. Polar Coordinate Form

      3:43
    • 9. Arguments

      4:11
    • 10. Euler's Formula

      4:07
    • 11. Exponential Form

      4:34
    • 12. Products and Powers in Exponential Form

      5:41
    • 13. Deriving Trig Formulae

      2:15
    • 14. Arguments of Products and Quotients

      5:05
    • 15. Roots of Complex Numbers

      5:18
    • 16. Summary

      1:16
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About This Class

In this course, you'll learn the fundamentals of complex numbers. We are going to cover the following topics:

  • Real and Imaginary Parts
  • Addition and Multiplication
  • Basic Algebraic Properties
  • Vector Representation
  • Modulus
  • Complex Conjugate
  • Polar Coordinate Form
  • Arguments
  • Euler's Formula
  • Exponential Form
  • Products and Powers in Exponential Form
  • Arguments of Products and Quotients
  • Roots of Complex Numbers

This primer will put you on a path to learning Complex Analysis.

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Dmitri N.

Quantitative Finance Professional

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Transcripts

1. Overview: Hi. I'm welcome to this first section off our complex number course. And in this section, we're going to just get ourselves started with using complex numbers. So what are we going to see in this section? Well, here are some of the things we're going to discuss now. First of all, we're going to talk about the addition and multiplication of complex numbers, a fairly important topic from which will derive subtraction and division as well. We'll also talk about some of the other useful algebraic. Probably ease off complex numbers and you'll actually see that complex numbers behave for the most part, just like ordinary numbers. So there is very little difference in terms off the kinds of operations that complex numbers support. Then we're going to talk about the geometric interpretation off complex numbers as vectors and the more July off those vectors and how they can be used. Then we're going to take a look at the complex contra, get a very simple idea which greatly simplifies Theo idea off dividing complex numbers, for example, and then we're going to jump into something really spectacular. We're going to look at the exponential form off complex numbers and how we can use that to actually let certain operations be a lot easier to evaluate, like, for example, taking a complex number to a particular power. They were going to take a look at products and powers in exponential form, So we're going to apply the exponential form to see what kind of benefits it brings. Will also take a look at arguments off products and quotients. Now an argument is once again something related to a geometric representation, all of a complex number. And finally, we're going to take a look at how to take a roots off complex numbers. 2. Real and Imaginary Parts: All right. So let's talk about how to actually set up the complex number system. So a complex number can be thought of not as one bad as two values I a pair of values. So typically you would write a complex numbers ed as X plus I y. So it's essentially composed off two parts. The ex part is the rial part. That's the part that you have in the rial number sistemas. Well, however, we also have the imaginary part the I Why? Where I is an imaginary unit. So we're gonna be using this exact notation throughout the entire course. And we're also going to be talking about complex numbers. Ask points, meaning that if you have a complex number which has X plus, I Y will be talking about it as a point x y on the so called complex plane. So just to reiterate the explanation once again, X is the rial part of the complex number and why is the imaginary part off the complex number and we actually have a notation for both of those things. So we have this r e notation meaning the Rio parts of the rial part of Zet is equal to X, and similarly, we have the imaginary part. So the imaginary part of Zed would be equal to why? So from this definition, we conclude that two complex numbers are equal if and only if they're real and imaginary parts are equal. Okay, so let's take a look at an example. So let's suppose that you have the complex numbers at being equal to three plus four I. So then the rial part off the complex number is equal to three. That's this part, and then the imaginary part, the part with the I in it is equal to four. That's this part. Of course, you can have numbers which are missing either the Rio or the imaginary part. For example, the Rio part off I, which is a purely complex number with no riel component, is equal to zero and similarly, the imaginary part of number 42 which has no complex component or no imaginary component. Rather, he is equal to zero as well 3. Addition and Multiplication: So now that we have some sort of a definition for a complex number, let's take a look at calculating the someone product of complex numbers and we'll begin with this some. So we're going to suppose that we have two complex numbers. That one is that too, with their respective real and imaginary components called X and Y respectively. And we're going to try and define this some. And really, this sum is actually quite easy because all you have to do is you write out the components . So you take the component of Zad one which is this part. So you sort of take these two values and then, of course, you take ah, these values. And what happens then, is you group the rial with the rial and the imaginary with the imaginary. Remember, you cannot really mix the two dimensions. All of a complex number. You cannot mix the Rio and the imaginary. So we take the rial parts, which are X one and X two, and we put them together. And then you take the imaginary parts. Why one and why to with an eye in front and you put them together and you therefore get a new complex number as the result. So the some off two complex numbers is typically a complex number where you have the rial components added together. Plus I times the imaginary components added together. So this is the sum, and it's fairly easy. However, the product of two complex numbers is a bit more complicated, so let's take a look at how to calculate that. So, first of all, as always, we write out the two numbers in round brackets, and then we multiply the terms so we multiply. X one by x two x one By I Y two you'll notice that what I do when I write the multiplication is instead of writing something like X, I Y two, I typically put the I either in front of the multiplication or at the very back. Let's just something that helps us read the number. So here I am, doing it in a kind of alphabetical order. So I x one y two. Now the interesting thing in this whole multiplication is the multiplication off i y one by i y two. Because this becomes I squared y one y two and remember, I squared is by definition. Negative one. So you get minus Why one? Why, too? And that's why we have a minus right here. So after we group the terms once again So we take the rial terms and put them together, and then we take the imaginary times and put them together. We have a more complicated expression. Now, here is something that's very interesting. If you take why one and why to to both be equal to zero, meaning that you turn zed one plus set to u Turn those numbers into purely riel numbers with no I component no imaginary component because this is, ah, the imaginary values. So you get rid of the imaginary values and you're left with the real part. Only then said one plus it too. From these above definitions from this definition and this definition. Zed one pleasant two is equal to x one plus x two and the same with multiplication. So if you take the why values off zero, then look what happens. You get rid of this because that's multiplied by a why you get rid of this so there's no imaginary component. Then you also get rid of this and you end up with X one x two. So you'll notice that if we start using real numbers under the complex number calculations , you'll still get the same values as if you were doing ordinary riel calculus. You would have x one plus x two and x one times x two And what this means is that the complex numbers are an extension off the rial number system. They behave with the same laws if you just let the imaginary components be equal to zero. So having acquired this functionality for multiplying two complex numbers, we can now see if some of our periods results are in fact, correct, because we know that we have this formula for a product. What we can do is instead of letting the complex parts the imaginary parts be equal to zero , we're going to let the Rio parts be called zero and will set. Both said One ends it to to have the value off I, which can be expanded a zero plus one. I. So this is the rial part once again, and this is the imaginary part you can think of those as you know, this is X one and next to both of them on this is why one and why to So why want and why two are equal to one, and the exes are equal to zero. And what you can do after that you do. This is you can substitute this back into our multiplication. And here is what you get. So you plug in the values off X one x two being equal to zero and y one y two being equal to one and you'll notice that most off the things vanish. So you get a one time zero here one time zero here. So there is no imaginary component and you get zero time zero here and you get minus one times one which is equal to minus one. So in other words, what we've just done using this ah, formula is we've proved the identity that we based this entire course on the identity that I squared because we said both off the numbers to I is in fact equal to negative one 4. Basic Algebraic Properties: Let's take a look at some more identities that we can come up with. We can determine the additive and multiplicity of identities meaning numbers which, when applied to the complex number, do not change it. And in the case of addition, that number is zero. And in the case of multiplication, that number is surprisingly enough is one. So you see that the identities in the case off complex numbers are actually really values. So you just add a zero here. Of course, you can treat the zero as zero plus zero high if you want to. But you don't have to take the imaginary component because, well, you you can acknowledge its absence. Or you can just say that we're taking a real value of zero, and the same goes here with a one. So a complex number multiplied by a one which is a real number, gives you that same complex numbers. So those are the additive and multiplication identities. And of course, we have the additive inverse. The additive inverse is of course, that number, which went added to the original number, gives you zero. So in this case, this is very obviously minus zed, which is minus X minus high. Why? So this essentially satisfies the qualities that bless minus that is equal to zero. And that's how we can actually introduce subtraction, because we can then say that Ah, the subtraction. So we can say that, said one minus. That, too, is the same as said one plus negative. That, too. So we have this definition off the additive inverse. Now the inverse actually happens to be unique because if you look at the definition so let's suppose we want to add a number two. It's inverse. Now we have to satisfy the condition that both the real and the imaginary parts cancel out . So the A has to cancel out with minus C, which means that the equality that a plus B I plus minus C minus d ie is equal to zero. That effectively implies that both A is equal to C and B is equal to d. Both of these things have to hold because remember the rial on the imaginary parts off complex numbers, they exist in kind of separate words. So we need two of these to be equal, and then we need be to be equal to D for the other parts to cancel out. Another thing that we can find in terms off complex numbers is the multiplication in verse , which is defined as that to the minus one. So for any non zero complex number, there exists such a multiplicity of inverse set to the minus one that multiplying the original number by this in verse gives you a rial number one. And this is called the multiplication of inverse. Now, finding that in verse is a bit difficult because we need to satisfy the following equation . We need the number which is X plus i y. We need that number multiplied by its inverse, which we've defined as a plus B I We need that to be equal to one now. In order to have this, we can expand out the brackets so we have a X plus b x I plus a I. Why I minus b y being equal toe one. And then, of course, we group the terms once again. So we take the rial terms, which is a X and minus B. Why we put them together and then we take the complex terms Well, the imaginary terms and we put them together and we have the following system. And this is actually a system of equation because we have to equation. We need a X minus b y to be equal toe one, because, remember, we want a one as a result, and we want the imaginary part to be zero. So be X plus a Y is equal to zero. So just here is a reminder once again of what we have. And we can now actually solve this system of equations in the following form. So I simply put it into the matrix form. And then you can apply any sort of rule that, you know, for solving systems of equations. So, applying Cramer's rule, for example, you have the following value for a So I simply get the determinants off both the Delta Matrix as well as the Matrix with the right hand side substituted. And that gets us a value off a, which is not exactly simple. You can see that the value here is X over X squared plus y squared, and in a similar fashion, you can solve it for be and putting it together. You'll see that the multiplication of inverse off a complex number, given the components. So we define Zed as X plus I y given these components all of a complex number, it's multiplication. Inverse is X over X squared plus y squared plus negative y over X squared plus y squared I . So that's the multiplication inverse. Not exactly a simple notation. And, of course, when Zed is equal to zero than the multiplicity of inverse or the division in the multiplication, inverse becomes zero. So the inverse of zero is unfortunately, undefined because we have this problem off. Plugging in the accent y components gives you a division by zero, so we cannot say anything about the multiplication inverse off a zero complex number. 5. Vector Representation: So far we've been treating complex numbers is just pairs of values, but you can start treating them as geometric objects because we can take a complex number and we can put it on a complex plane because it has two coordinates, the Rio coordinate and the imaginary coordinate and what we can do is not just treated as a point, but actually draw a vector from the origin to each of the complex points. So here is an example. Let's suppose that you are trying to display the complex numbers ed having the value off five plus for I. Now what you can do is you can plot the accent y coordinates where acts is the rial number line, and why is the imaginary numbers And then, of course, the 0.5 plus four. I has the coordinate five along the real axis and has coordinate for along the imaginary axis, and so you can put a point in here. But in addition to just having this as a point, you can also draw a vector from the origin to this point and then treat this point as a vector. And here is a different point, minus to plus six I you notice that here were minus two. So where? On the negative side, Off the rial number line. And we have a slightly larger value on the imaginary light. So once again, you can treated either as a point minus 26 or you can treat it as a vector, going from zero to minus 26 Now, with this in mind, we can have a geometric interpretation off the some and the product, which means that we can treat the some, for example, off two complex numbers that one. And that, too, as a sum of vectors. And if you add vectors, you end up with typically the trapeze Imu, which looks something like this. So we for adding the two vectors said one ends at two. Then you simply add one to another by attaching it to send. So here I can have said one, followed by this vectors. That, too, or, in other words, I can have that too, followed by a vectors that one. Remember, they're interchangeable, and we end up with a trapeze iam and the diagonal of entropy. Zi, um, happens to be the some Is that the one place that too now the product that ones. That, too, is somewhat different, and I want you to pay attention here to the differences between this kind of product and the kind of products that we have in ordinary factors. So let's remind ourselves off what the product actually is. First of all, a product is typically a complex number, which means that if that one ends it to our complex numbers, then Zed ones. That, too, is typically a complex number. And what this means is that in the geometric interpretation, it's still a vector. It's still a vector, which has two points X and Y, and this is different to a vector product or a scale of product. Now let me show you why it's different to a scale of product if you have two vectors. If you have a B and C D in the rial number system than their product happens to be a scaler value, their products happens to be a C plus B D, which can be a scale of value like 42 for example. The difference with complex numbers is that this product still gives you a two dimensional result. It's still a vector, so we can state that the product of two complex numbers is definitely not a scale of product now. The other observation that we can make is that the multiplication off two complex numbers still happens to lie in the same plane as both off the numbers being multiplied. This is also very interesting. If we multiply two complex numbers, which exists on the complex plane, which is sometimes referred to as this, then their product is still on the complex plane, which means that they don't escape the plane into some third dimension. Now, if you remember your linear algebra and if you remember a vector product off two vectors, you have a vector like this in the vector like this. Then there vector product, the sort of a cross product be there. Vector product would actually go into the third dimension perpendicular to both off the both of the original vectors. And what this means and why this is relevant is that in our understanding, the multiplication off two complex numbers said ones. That, too, is neither s Keylor nor a vector product. So even though we're doing this geometric interpretation, their product cannot be interpreted as either a scale, a product or a vector product. In our familiar linear algebra terms, 6. Modulus: having introduced the definition off complex numbers as vectors, we can now talk about the modular off a complex number which corresponds to the length off the vector that's representing this number. In other words, is the distance to origin. So if you have a point, let's say you have a point two plus two I. So that would have a to coordinate on the rial access and a to coordinate on the imaginary axis so the point would be here and the vector would be here. So the module ISS, which is denoted as the modular of Zed, happens to be the length all of this vector. It's the distance from origin to the actual point eso, given that that is equal to X. Plus, I y. Here is the definition off the modular so fairly obvious stuff, and this is known as the modular off the complex numbers ed. And the interesting thing about the module is is that it's really the complex counterpart to a really numbers absolute value. Now, remember, in the rial numbers, you basically have a really number line, so it has a zero somewhere, and then the model is often ordinary number or the absolute value of unordinary number is the distance from zero. So if you have a number minus three than this distance, which is the absolute value, off minus three is equal to three, because that's how many steps you need to take to get from minus 3 to 0 now with complex numbers. This is particularly important because this is really the only way that you can compare complex numbers because the notation said one less than that to has no meaning. The comparison between two numbers has meaning. When you compare their distance to the zero, for example, so we can say that three is greater than two because it's further away from zero. But with complex numbers, it's a two dimensional picture. You can't really compare this point to this point and say that this one is bigger than this one, for example, but you can compare the distance to zero so you can compare this distance to this distance . And that's why the only way off comparing complex numbers is by comparing their more July so we can post some interesting findings with relation to the module is first of all, we can define a distance between two complex numbers, and that's rather easy. That's simply the distance between the points. And if we have the X and Y coordinates of those points, then finding the distance is rather easy. We can also define an equation where you have some points at zero being subtracted from a complex number. And then if we say that the distance between these two numbers is some radius are than what we've essentially done with this definition is with defined a circle. So we've defined a circle of radius R centered at some 0.0. So here is the Radius, and this equation effectively defines the entire set of points that I've drawn Now. We can also express the complex number in terms off the real and imaginary components in the following fashion. So you'll notice that if we just take the ah sort of magnitude in the accent y direction than the magnitude happens to be the real and imaginary components of a complex number. So that is essentially the X distance, and this is the wide distance from origin. And this way, what we can do is we can say that the ah modular off the complex number squared is the same as the rial component squared, plus the imaginary component square. And finally, we also have the triangle inequality, which kind of mirrors what we know in ordinary geometry, that is, that the module is off the some off two complex numbers is less than or equal to the some off them on July. 7. Complex Conjugate: the complex conjugal is a really useful to for calculating some things like, for example, doing division in complex numbers. So given that we have a number zed defined as X plus I y the complex contra git off this number Zed, which is denoted by said bar, is simply X minus. I y so a very simple idea. Now the complex conjugated obviously represents the point where the X coordinate is the same, but the white corn that is different. So if we were to draw it and if this, for example, is the point X plus I y, then the complex conjugal is simply the point reflected along the X axis. So it's the same distance from origin. Obviously, there is a different angle here, but it's essentially the same except that it's negative. I y here, OK, so we can determine some relationships between ah, the ah Contra git and the number. For example, Taking the conjugal twice or reflecting twice along the x axis gets you back to where you started. Also, the module is off the conjugal. It is the same as the model is off the number itself, because if you look at the image here, you'll see that the distance is exactly the same. It's just reflected so we can find additional relationships. Like, for example, the conjugal off the some off two complex numbers is the same of some of the congregates, and similarly, the conjugal off the product is the same as the product of congregates. I am posting these here without any proof, but you're welcome to right out the results by simply substituting ah, the appropriate value. So doing the multiplication terms of x one y one x two white and writing this stuff out. But I'm just putting this here as they are. In addition, what we can do is we can state that Ah, the some. So the some off the number and it's conjugal can be written out like the following. So we have the complex numbers add, which is X plus I y added to its conjugal. What this implies is that the imaginary parts are going to cancel out blast high wind minus I. Why they cancel out you end up with two acts, and similarly, if you do that, miners that bar, you'll just end up with two high y. And from this What you can determine is you can determine the following relationships, so the rial component off a complex number is that number. Plus it's conjugal, divided by two, and similarly, the imaginary component off that number is that number minus is conjugal, divided by two. Also, you'll find that the number multiplied by its conjugal is equal to the modular squared, and this is a critical part. So this isn't something that we should ignore, because one of the things it helps us do is it helps us divide one complex number by another. Now division is typically difficult, but with this little trick, this becomes really easy. Here is why here's what you do. So let's suppose you want to divide a complex numbers at one by a complex numbers at two. Now you could be having something like two plus three. I divided by four plus five. It's not immediately obvious how to do this, so what you do is you multiply the top and bottom parts of the fraction by the complex con . Trick it off the bottom and look what happens here. You multiply this by the complex conjugated full four plus five I, which gets you four minus five. I multiplied by four minus five. I now this top part this multiplication weaken. Do we know how to multiply complex numbers? We have a formula, for it is going to work out fine. The critical part is this part. This part is going to be equal to the modular us off four plus five. I squared and this is a real number. This is a really number. It's not a complex number. So you end up with some complex number on top A plus B. I divided by a real number and because it's divided by a real number, you can rewrite it into the appropriate part so you would have a over Ah said squared. Plus B over That's squared I so you can see that the division, when you have one complex number divided by another, can be really easily figured out by simply multiplying the top and bottom by the contra. Get off the denominator, and then the bottom part becomes a real number. That's how you divide complex numbers 8. Polar Coordinate Form: So far, we've been talking about complex numbers in purely Cartesian terms, where the complex number has X and Y coordinates or the vector ends up at a point x y. So we already know this, but now we can introduce a second representation off complex numbers, which is also geometric. But we're going to represent the complex number or the vector in terms off the length of the vector and the angle of inclination. So we're gonna have this formula, and I'll explain it in just a moment. But one constraint I want to put on this formula is that the distance from origin off the vector is always assumed to be positive. A zero. It cannot be negative. So here is the illustration, and let's try to go through this slowly. So we have a complex number four plus three I and we already know that we can plot this on their Cartesian coordinates by taking X to have the value for why having the value. Three. So here is our point, and that is one way you can determine the point just by stating that the point has coordinates a four and three, and that's the Cartesian system. However, in polar coordinate form, what you do is instead of defining the ah X and Y coordinates, you define to other things. You define the angle of inclination which will call Fada and you define the length off the vector. So you define this distance and of course, we already know that this distance happens to be. The module is off the number. So here, if we want to take four plus three, I we can easily find the module is because that's three squared plus four squared and the square root so the length is equal to five and faith is the arc tangent off 3/4. So once again, if we were to now take this definition if we were to take this information as the starting point and we wanted to find the components off the the actual ah at the X and Y, the Cartesian components that this distance from 0 to 4 would be defined as are caused data . So the rial part off this number would be our times cause failure and then the imaginary part would be our times. Sign Fator And let's not forget that these are imaginary units so this is always multiplied by I. So you would have I multiplied by our scientific and then we can factor out. The are so we have our times cause data. Plus I, Scient, Ada and that's exactly what we have right here. So this is where we have our factored out. And then we have cost eight a plus. I sine tade. So what we've now done effectively is we've introduced another notation another way off, expressing complex numbers that instead of using Cartesian coordinates Axum Why it's using the polar coordinates. R, which is the distance, which is The module is off the complex number and theater, which is the angle, and you'll notice that the angle is calculated in a counterclockwise fashion starting from the axe axis, and it's calculated in radiance typically. So, for example, this angle from here to here would be pi over two. Or this angle, for example, would be pie. And if you want to take this angle, for example, this would be negative pi over two 9. Arguments: So now we're going to introduce the notion off the argument, and that's a very simple notion. So when we define a complex number in the form that we have just seen, we have this angle theta and the angle theta happens to be called the argument off the complex numbers ed, and the way we write it is our exit is equal to theta. But the thing about the stada is you can really reach a complex point by rotating as much as you want, which means that if you want to reach this point, for example, you might think that there is a new angle going like this that helps you reach this vector helps you reach the actual point. However, nothing really prevents you from doing a full circle a full two pi circle and then going like this. So the number off angles which get you this vector is actually infinite. It's an infinite number of angles. You can rotate as much as you want and the problem with this the problem with the fact that we can rotate as much as we want is that we need to introduce some other notion off a unique angle. And this unique angle is called the principle argument, and you'll notice that, unlike the ordinary argument with a small A, the principal argument is written with a capital A. So the principal argument off a complex number is a unique value. Capital Fada such that it gets you the right point, so it gives you the right angle. But it's constrained, so it always has to lie between negative pie and pie. So what this means is that if you have a point here, you can no longer do a full circle because doing a full circle will give you two pi and two pi is too big. Remember they that Capitol Theatre has to be less than or equal to pi, so it cannot be that so. It kind of constrains theano girl. And then, of course, if we put those two together, we have the following expression. So the argument off a complex number is its principal argument, plus as many rotations as you want to do because two pi gives you a single rotation and multiplying it by end where and is a whole number that gives you any number of rotations in any direction. because you can set and to be equal to minus for the two, for example, and you'll be doing lots and lots of Coke wise rotations before you reach that point. So just to reiterate the principal argument is a unique value, which gives you the angle. And the argument itself can be virtually any principle argument, plus any number of rotations. Okay, so let's take a look at an example off finding the argument. So let's suppose we have a complex number said, being defined as minus one minus I so we can do a diagram like the one I have here, and then we can see that the principal argument happens to B minus 3/4. Spy. Now you might be wondering, Well, why Onley this argument? Why not go the other way around? So we've taken this arrow here to be the principal argument. But why No, the red one. And the reason for that is that the red argument is actually five pi over four, and that's greater than pie. And the problem is that we have this constrained on the principle argument. The principal argument has to be between negative pie and pie, and we have something greater than pie. So obviously we cannot claim that this larger angle is the angle that we're looking for. And in terms off the argument itself, we can say that the argument is the principal argument, minus 3/4 Spy plus two pi A. And of course here, if you set and to be equal to negative one, for example, you would get a particular ah value with an extra rotation or if you set, and with the value being equal to one, you will get exactly this pi over four value five pi over four. However, the principal argument is unique, so there is no other principal argument for this complex number. 10. Euler's Formula: we're now going to derive a very famous finding by Oiler, which is called Oil is Formula. And in order to derive this finding, we're going to talk about the McLaurin Siris for some off the well known trig functions. So if you remember your serious expansions for sign and cause they are as follows. So you have the sign expansion being X minus X cubed over three plus X to the 5/5 factorial minus X to the seven of 1/7 pictorial and so on. So you noticed that the signs alternate. It's minus blast, minus blast and so on. And we have effectively the ah value here and the value here being increased by two at each step. So it's 357 9 11 and so on, and the situation for co sign is similar. So we have one minus X squared of a two factorial plastics to the four all the four factorial minus X to the six of a six pictorial. So it's once again increasing by two, and the sign alternates between minus and plus. Now let's also remind ourselves off the Siri's expansion for the exponents. So the serious expansion for the exponents is one plus the number plus the numbers quite over to a pictorial. Plus the numbers cubed over three factorial and so on and so forth. Now we're going to do a trick. What we're going to do is we're going to take this value of Zed, and instead of treating it as a real number, we're going to plug a complex part in it. We're going to define Zed to be X multiplied by the imaginary unit, and we'll take this definition and stick it back into this equation. So we'll take the i X definition. It will plug it back into the serious expansion for the exponents. So what we're going to get is the following we have of the serious expansion looking like this. So we plug in the values first of all, so we have i X and here we have I X. And here we have I accent here and now we remind ourselves off the power's off. I What does it mean to square high? Well, we know by definition, that I squared is equal to negative one. So I cubed then is equal to I squared times I and we know that I squared is negative one, so that becomes negative high. However, if you do I to the four, for example, that is negative. I multiplied by I, which is equal to one. So you have this nice, interesting alternation And if we write it out, we'll get something like the following. Get one plus i x Then I vanishes minus I X cubed over three. I vanishes again. There is no I hear plus i x to the 5/5 pictorial. Now you're going to find something interesting here, let me highlight the terms of which do not have an imaginary component. So it's this term, this term, this term and so on. So these terms do not have an imaginary component. Now, if you look at this term this term and this term, you'll see that they are identical. They are identical. So it's the co sign expansion. And similarly, if you take the imaginary terms which are I X minus, I X cubed over three factorial plus i x to the five of five factorial. And if you take out the I, you'll see that it's the expansion for signs. So here is the X that we have right here. Here is X cubed of a three factorial Here. This Ah, here is extra. The five of five factorial here it is. So what we can do is we can factor out the eye and rewrite the expansion in the following fashion. So it's the whole terms, the rial terms, plus I times the imaginary terms. And then, of course, we plug in the fact that this whole part, this front part, happens to be the expansion for co sign and the part multiplied by I happens to be the expansion for sign and we end up with a rather interesting finding. We end up with a finding that each of the I ax is equal to cause X plus I sine X and this happens to be oil this formula. 11. Exponential Form: so armed with oil ist formula, we can now derive a very famous equality. So remember that we have this definition for a complex number. So zed in polar coordinates is equal to our coast data plus i r sine theta. But we now have a new definition from oil is formula. We know that coast data, plus I sine theta is actually the exponents. So we can simplify this formula by rewriting this part as ah, the exponents here. So what we can derive from this is the following notation. So the complex number zed can now be represented as its magnitude multiply by the exponents off I multiplied by its argument. And here is where things get interesting. What you can do is you can start plugging in values directly into this equality. So, for example, if you plug in r equals to one and theta equals two pi, you get the following diagram. So let's just explain what's going on here. So we have a unit circle and the angle is pie. This is the angle. So we rotate by pike counterclockwise. We end up at a point negative one. So what this means in practical terms, is the pie that gets plugged into here and the one that gets plugged into here gives us the quality eats of the I pi equals negative one. This is also sometimes reading as each of the eye pie plus one equals zero, and that is a very famous equality. But if you want to, you can derive other similar equalities as well. So, for example, if you rotate the angle not by pie but by pi over two, for example, you end up at this point So if you rotate just by by over two, you'll end up at this point which is, of course, I So it gives us the following equality E to the I pi over two. So now it's by over two radiance that gives you the value off I. And similarly, for example, if you rotate a full circle so if you do a full circle from this point you end up at this point and this is a 0.0.0.1 with no imaginary components. So eat of the two pi I is equal to one. Okay, so we now have this exponential form for complex numbers, A really useful tool. So if you have a complex numbers ed being equal to negative one. Minus I. You can represent it in this form by festival, calculating the components, calculating the are and the theater. And this is done rather easily because the are is simply the module is so to calculate, the module is off something. You essentially do a square root off the Rio part, plus the imaginary part. So the real part is negative one. So you do negative one squared plus and then you look at the imaginary part, which is also negative one. So you have negative one square, so that becomes the square root of two. And that is your magnitudes, that the length of the vector essentially. And then, of course, the argument fada is calculated as the angle that you need to rotate. And if you do applaud, if you just, you know, draw the drill, the actual point with the sort of X and y coordinates, you'll see that negative one minus. I is right here, so it's exactly ah, it's exactly here. So the angle that you need to turn to get to this point is 3/4 off the pie clockwise, so that's minus three pi over for. However, remember, the argument is really that number plus any number of rotations you could do any number of rotation. So instead of doing it like this, you can do it like this bow that we go. So that adds an extra two pi. So we define the argument or the data as minus three pile of four plus to buy and where and is a whole number once again. So our definition off this complex number minus one minus I in exponential form looks like this. So it's the magnitude times the exponents off I multiplied by Fada and theta is exactly this part. By the way, if you're not familiar with the notation here when we write e till the something and that something is very large then instead, what we do is we write exp and we just ride that something very large in line as opposed to in superscript 12. Products and Powers in Exponential Form: so having this exponential form, we can now calculate the products and powers off complex numbers in this exponential form, and you'll see why exactly we did this. So let's imagine that complex numbers have no riel component. Let's suppose that the they all exist on a unit circle, in this case, the product of two complex numbers. Let's say you have just the theater components of each of the eye fate a one each of the eye theater to Well, you can add up the theaters. Obviously, you can add them up, and therefore you have each of the I multiply by fate, a one plus data to If you now add the magnitudes as well, you get the following. So the product of two complex numbers in exponential form is the product off the magnitudes . There really values multiplied by the exponent of I times the sum off their arguments. So and you can you can obviously see that this this leads us to a formula with there to the end. We'll get to that in just a moment. But first of all, let's talk about the inverse once again. So the multiplicity of inverse off a complex number set to the negative one is essentially one divided by Zet. And once you have one divided by said you can take both of the one and the zed and you can represent them in this exponential notation. So one is really one multiplied by eats of the eye. Time zero and, uh, the number said is by definition, are multiplied by Ito, the I Fada. So then what we can do is we can take the one over r and that's a whole number. So we can just Well, it's ah rial number. So we can just take it out in a za fraction here. And then we're left with each of the eye off zero minus data. So we end up with one over, are multiplied by eats of the minus. I faded, so that's a need find another need find is that power Is that to the end, Like I promised says that to the end is obviously a collection off. These are multiplication multiplied by as many fate as being added up is needed. So if you have not won in two, but if you have 123 all the way up to end you'll have ah R to the power of and here because there's gonna be as many or as being multiplied. And here it's in addition. So if you're multiplying Fada and Times Nets end times data, obviously, and that's exactly what we have here and what we have here. So this is our definition for taking a complex number to a particular power because we simply take the the appropriate components and sort of bunch them up together, so to speak. And, of course, if you have a unit circle, if you have are being equal to one than a set to the end is equal to E to the high Anthea, That's another interesting finding. Now let's take a look at taking the power of a complex number. So suppose you want the squared off three plus I to the power of seven. You don't really want to multiply out all those terms because you could certainly write it as Route three. Plus, I'm multiplied by three plus I and then try to write it out this way. But it is time consuming, and we don't need this anymore because now we have the exponential form. So with the exponential form. What you can do is you can write, are as follows. So festival you get the route three plus are into exponential form. So you calculate the magnitude that's essentially Route three squared plus one square that gives you a value of two. You calculate the argument, which is Thea Arc Tangent off one of her three, which is pi over six. And then you have the are in the Fada so you can get your get your complex number in this R E to the I fade before and so this is exactly what we do. So, first of all, we take the complex number and we put this in the form to which is it's ah ah, it's magnitude times e to the pie I over six. Here's the pi over six. It goes in here to the power of seven. And then, of course, we can start treating it as just an expression where two to the power of seven can be factored out, for example, multiplied by each of the seven pi over six. And we can sort of stopped at this point, but we can go even further. Let's suppose we want the expression in terms off the original value. What we can do is we can note that to to the seven can actually be divided into two through the six multiplied by two and similarly eats of the 76 pie I you can essentially subtract or you can split it into two multiplication Xas. Well, so here what we can do is we can multiply heats of the pie. I buy E to the pie high over six because this is one pie I and this is a six. So one and the sixth gives you 76 That's how we split those two up. And now what we can do is we can take this with this and this with this and that way we end up with the falling multiplication is we end up with two to the six times E to the pie I, which by definition is minus 64 because the magnitude is to to the six and pie gets you a negative one. Remember, we have this famous result that eats of the hi I is equal to negative one. So that's 64 multiplied by minus one, which gets U minus 64 then the other side to eat to the Pi over six. We've seen this before because that's the original number. So that's the original number of that's with three plus I. So the answer is minus 64 plus Route three plus one. 13. Deriving Trig Formulae: one very interesting consequence off working with these complex number qualities. And the exponential form is that now that we have the exponential form off our to the end, what we can do is we can start deriving trigonometry formula. So let's take a look at how this is done. So we're going to make the assumption that we're choosing the power, the and value when takings at to the power and will choose the value off to so an is equal to two and in addition will set are to be equal to one. So we're working with the unit circle, so this whole thing becomes a one, and we're going to ignore it. So for our purposes, is it to the end? Or is that squared is going to be equal to e times to I fada. In other words, the angle here is tooth ada And because the angles to fade, if we go back into some of our earlier notes, we know that this is actually equal to cause Tooth Ada, plus I sine to waiter, right? I mean, we've had this before ready the angle to Fada being here in here. So we end up with a really interesting equality. So if you take Zed to the power of N where N is equal to two, so said Squared, you can write it like this. So it's called State A plus I sine day to remember we ignore they are completely because R is equal to one. So any power off our is equal to one. So we have called stated. Plus I sine theta square because we have a square here. But on the other hand, we've just derived this part. We've just arrived the part where it's cost to fade A plus I sine two theta and that's really interesting because what we can do is we can take the left side, we can take this entire side and we can expand the brackets. And if we do, we end up with the following notation. And then, of course, we can now equate the real and the imaginary parts off this thing and this thing because this has really an imaginary parts and this is real and imaginary parts. If we equate those, we get the following to formula, which should be really familiar to. So these are the call sign and sign formula for a double angle 14. Arguments of Products and Quotients: All right, now we're going to talk about the arguments off products of complex numbers, So some of this stuff should be really obvious. So we've we have determined that the product of two complex numbers is the product of the magnitudes times the exponents of i times this some off their arguments than what we can determine from that. Is that the argument off the product? Is this some off the arguments? Because I noticed there is a plus and here, So that's should be an indicator that if you're multiplying products, you're getting the argument, which is the sum. And similarly, if you're dividing one complex number by another than the argument of the division is the same as subtracting the arguments. So we can actually illustrate this with a example where we get to actually draw something. So, for example, let's suppose that zed one is equal to negative one. So I'll actually draw this on a diagram as before. So said one is equal to negative one, which is here. So said one is exactly this point and set two is equal to I So set to is this point Okay, now, on the one hand, we can calculate the argument of the product. And if we multiply negative one by I, we get minus I. And of course, the argument of negative I, which you know, is here somewhere. The argument is equal to minus pi over to Hopefully, that's obvious, but only are the hand we can calculate their separate arguments. And here you notice I started using the capital? A. So I'm talking about principal arguments, not about rotating a zillion times before you reach the point. So the principal arguments of said one ends at two are as follows. So said one is here that obviously requires 1/2 circle. So the angle off pyre a Deion's. And to get to this point, you need pi over two. So pi over, too, for the negative, I So that's what we have. We have a pie and pie over to as the argument. So if you add them up, however, you'll notice that they are different. They are not identical because you have the The argument of the product is negative pi over two. But the art of the some of the arguments is three by over two three by over two, so It's really weird that we claim here on the left that the argument is the sum and here it's not. But notice the capital a here because the ordinary argument allows you to rotate as much as possible. And if you rotate minus by over two, just enough. You end up with three pi over two because three pi over two is greater than pie. That's too big. So if you have this value, then this value is not a principal argument of anything because it's greater than five. So essentially the take away from this is that the argument, the principal argument off said one overs that two plus the value of two pi. So if you take minus by over two and you are to pie, you do in fact end up with three pi over two, or you can start writing them as ordinary arguments. So the ordinary argument off the product is equal to this some of the arguments, because remember, with the ordinary argument, you can rotate at any point that you want. Now let's take a look at yet another example, and in this example, what we're going to do is we're going to Try and Khalkhali the principal arguments off the following division. We have minus two. Divided what? One plus the square root off three I. So how do we do this? Well, we note that the argument off this number is the argument off the numerator minus the argument off the denominator. Because, remember, we have this formula from two slides back where the argument off a over B is the argument of a minus the argument of Be So that's exactly what we didn't hear. We're taking the argument off the numerator, and we're subtracting the argument off the denominator, which, by the way, should be, ah, the root root of three I Sorry about that. So since the argument of minus two, which is a negative numbers pie just once again just to reiterate minus two is here. So the angle is obviously pie and the argument of one plus route three I is equal to pi over three. Yes, I used math lab to calculated. Then we can determine the following. So we have one value off the argument of zed, which is pi minus pi over three. So we take this part minus this part, and we get pi minus by over three. So that's 2/3 spy, and this just happens to fit the range off Arg. Because remember, we have the principal argument has to be between minus pie and pie, but 2/3 pi fit exactly between this range. So this means that the answer that we have may as well be specified as our exited with a capital A as well. 15. Roots of Complex Numbers: this is going to be a bit more complicated, but essentially we're going to take two complex numbers, said one and set to. And first of all, let's remind ourselves off the equality criterion. How are these two numbers considered equal? Whether considered equal if their magnitudes are equal and their angles are equal with the tolerance off? Two Pi K, where K is a whole number because, remember, you can rotate as much as possible. Unless, of course, we're talking about principal arguments, in which case you kind of lose this fax ability. Now, using this definition, we're going to make another introduction. So we're going to assume that we have some complex numbers. Ed, which is defined an exponential form, is our times e to the i theater, which happens to be the 10th root off zero. So we have that zero, and we want to take some and threw it like the cubic root of it, for example. Now an exponential form is going to look something like this. We have zed and remembers that is actually the route. So zed to the power off an So are to the end e to the I. Anthea is equal to If we defines that zero as through our zero and Fada zero, then this is the definition we have. So taking Zed to the power of n we get the left hand side and that is equal to the right hand side. The one with our zero and fate is here. Okay, so so far, so good right now, given our equality criteria that we've defined meaning this stuff, what weaken state is that the parts have to be equal meaning that are to the end has to be equal to r zero. And Anthea has to be cool to fade a zero. So this part has to be equal to this part with the tolerance of to bike A Because once again, we have this rotation thing going on. In other words, for the ah kind of ah riel part if we talk about our to the end and are zero, we know that our is the end through of our zero. So in terms off the rial part, the rial components, everything is ok if we if we know that if we know the rial component off the number of want to take some root off then we can easily determine Ah, the rial component off the actual route. That's that's fine. And similarly, for the fade away, we have the following. So we have the data. If we kind of try to factor out this data here by dividing everything by end so that we can get rid of this end. Here we have data equals Theda zero plus two pi ke over end and then we can sort of split it down the middle. So we have the part with Ada zero over and and we have to pike a over it. Okay, so really, the n threw off some numbers that zero is equal to the following. We have the rial part which remember, we kind of found here. That's the easy part. Plus the exponents off. I times this expression this expression up above where k is once again, a whole number. We can actually apply this formula for finding the end through that something by solving an equation like for examples that cube equals negative one. So you might think that there's just one route that that is equal to minus wine, and we should all go home because there are no other routes, but in turns out that this equation actually has three roots and you've only guest one of them. So first of all, since we want the cube root of negative one, we define negative one in exponential form. So that zero, which happens to be negative one, by the way, is equal to eat of the pie. I remember we have this famous each of the hype. I hope I I being called to negative one. So that's what are complex number actually is, or the complex number representation off negative one is. And then we apply our formula and it looks something like this. So remember, we have no magnitude. We have a unit circle. But we had, ah, the Fada over N plus two by K over end and remember in this case and is equal to three. So that's why you have a three here and the three here. So now we have this solution, and the range of K values goes from 02 and minus one, of course, and minus one. This three minus one. And that's equal to two. Which is why the limiting condition is to here so we can start plugging in these values off K and seeing what expression we have. So if you're plugging cake zero, that's easy because you have each of the i times pi over three and that's equal to cost by over three. Plus I sine pi over three, which is equal to 1/2 plus Route 3/2 I. So that's one solution. Then, of course, we plug in K equals one, and in this case we get the square root of zed. And if you actually plugging cake whilst one, what you get is you get 1/3 pie plus 2/3 pi. So you get a pie. So you get each of the pie I, which is equal to negative one. That's the intuitive route, because minus one times minus one times minus one does, in fact equal minus one. So that's the route that you might have guessed. And then, of course, four k over to you get similar thing to with K over cankles +20 in that you have the exponents of five by over three, which is equal to cause bio five pi over three. Plus I sine pi over three, which is equal to 1/2 minus the square it off. 3/2 I. So this way we found all the three routes off the equations. That cube equals negative one. 16. Summary: So let's try to actually summarize what we've learned in this entire section. First of all, we saw complex numbers and the fact that they can be represented, as actually two pairs of numbers the rial component and the imaginary component which we can then treat as either a point or a vector. We saw that complex numbers retain all the futures or real numbers, so things like associative ity, community and so on. Essentially, they are an extension off the rial number system. We saw addition and subtraction and the treatment off complex numbers as just ordinary numbers with real and imaginary part. Separately, we saw the complex conjugal which was very useful for a division we sold. The polar form are times cost eight a plus. I, Scient, Ada. Where are is the module is offset and theta is the argument of zed. And we also saw, for example, that weaken jump from polar to exponential form, using the wonderful oil ist formula, this one and then we saw that this makes it very easy to take. For example, the power's off complex numbers. We also saw that roots of complex numbers can be found according to this formula, and that was our latest finding