 I'm Zor. Welcome to Unizor Education. I'd like to present the problem on the topic of complex numbers. It will be problem number two, actually. Here it is. It's related to graphical, geometrical representation of complex numbers on the plane. So let's consider you have a unit circle and a complex number on it. a plus vi, and it's on the unit circle. It means that a squared plus b squared is equal to one. This is a squared. This is a, and this is b. So a squared plus b squared is equal to one. So that's what it means that it's on the unit circle. Now, let's assume I'm considering a square of this number. The problem which I would like to present is the following. The square of this number is also on the unit circle, and these two angles are exactly the same. So basically to square the number which has a representation sitting on the unit circle, you just have to basically use the same angle which it has with an x-axis and just double it basically. So the angle between this radius and the x-circle is exactly double from the angle between this radius and the x-circle. So this point is a plus vi squared. So that's the problem. Two parts begin. One part is that the square belongs to the same unit circle. Second part is that the angle doubles. Alright, let's prove the first part. That's the easy. Since the distance between two different points which represent complex numbers is basically a modular of their difference. Well, this is zero, so it's a modular of the number itself. Let's find what's the modular of a plus vi. Obviously, we'll use the fact that a squared plus b squared is equal to one. So what's the modular of a plus vi? A plus vi squared. We need to know the modular of this number. Well, let's just do the calculations. Square is a squared plus b squared i squared and i squared is minus one. So it's minus b squared according to the rules of multiplication of complex numbers. So this is the real part of our new complex number. And our imaginary part will be ab to abi. Okay? What's the modular of this? The modular is the square of the real part plus the square of the imaginary part. Right? Well, this is not the modular. This is actually a square of the modular. So if I will prove that the square of the modular is the same one, that's the same as modular is equal to one. But now, what is this expression? Well, that's very easy. It's a to the fourth minus two a squared b squared plus b to the fourth plus four a squared b squared which is equal to a to the fourth minus two a squared b squared plus four a squared b squared. So it's plus two a squared b squared plus b to the fourth. And obviously you have recognized that this is a squared plus b squared which is equal to one because a squared plus b squared is equal to one. That's the condition, initial condition that we were using our initial point. So proof that the square of our complex number which belongs to the unit circle also belongs to the unit circle. That's easy. Okay? That's the first part. I want to explicitly ask you to press the pause button before I start solving, proving this particular thing. Well, I forgot actually. Now you can press it because if you want, the most important part of this problem is number two actually, that the angle is doubled. So before I start this continuation of this problem, solving this problem, those of you who would like to try their own and I do suggest you do that, press the pause button and think about this yourself. But I will continue now with proving the second part of this problem that the angle doubles when I'm squaring a complex number. This is a little bit more involved but anyway, here it goes. I hope you did press the pause button. Okay. Let's consider these two segments. This, by the way, is one because it's a unit circle. This is also one. Now, obviously, since we are on the unit circle, so all these right radias are exactly the same and equal to one, if I will prove that this segment has the same length as this segment, that will prove that my angles are the same. So the total angle from x-axis to my new square complex number is double. So the only thing which I would like to prove is that this distance is equal to this distance. Now, as you remember, the distance between two points on the plane can be actually solved using called algebraic methods of complex numbers. That's the beauty of it. So I will use the complex number which represents this point, this point, and this point. I'll take their complex numbers difference and the lengths of the corresponding segments will be a modular of differences. All right. So the only thing which I have to do is basically just do a bunch of tedious calculations and I do have to tell that this is a little bit lengthy and tedious. It's simple. It's just plain multiplication and addition and subtraction of numbers, but well, you've got to do it and that would prove our problem, our theorem. All right. So first of all, let me express this a plus bi square as a more traditional complex number, which basically I was using before. It's a square minus b square. That's the real part and complex and a imaginary part is this. That's exactly what I was using to prove that this segment is equal to one. I just took the module of this and turned out to be a square plus b square squared, which is one. Okay. So this is traditional notation for this particular point. Now, at this point, we already have this notation a plus bi. How about this point? Well, this point obviously is real part is one. Imaginary part is zero. That's what we have for this point. So the distance from this to this should be equal to distance from this to this using a square plus b square is equal to one as a condition. Well, let's just put all these four conditions again on the board so I will have some free space. So a square plus b square is equal to one. That's one of the conditions. Now, my three points are point number one is a square minus b square plus two a bi. My second point is a plus bi. And my third point is one plus zero. From this time on, we do not really need any geometry. We will just use pure algebra. Okay. Now, let's frame this as a condition. Now, the distance from this to this. Well, you have to basically make this difference which is a square minus b square minus a as a real part. And plus two a bi minus b as imaginary part. That's one number. That's the difference between these guys. Now, the difference between these guys is a minus one and b minus zero i. So it's a minus one. That's the real part. And b minus zero. Well, that's just a b. Now, the parentheses i. That's the difference between these points. So what do I have to prove? That the module of this guy is equal to the module of this guy. Right. Okay. Now, we will wipe out again what we don't really need. With that, we need this condition. And all we need to do right now is what's the module of this calculated by square of this plus square of this? What's the module of this? Square of this plus square of this. And compare them, they should be exactly the same. All right. So this is my a, and this is my b. Two complex numbers. Module of a, I'll use two vertical bars just to signify the module. It's equal to, I will probably put this b somewhere here. Module of a, square of this plus square of this. Yeah, that's tedious, but anyway. a square minus b square minus a times a square minus b square minus a. That's the square of this guy. Now, square of that is, well, I can do it right now. It's four a square b square minus four a b square plus b square. Square of this number, which is this, minus double this times this, which is this, and plus this. Okay. Equal. Now, we have to multiply these. All right. Let's first multiply a square by each of those guys. We will have a to the fourth minus a square b square minus a cube. Now, minus b square times all these guys minus a square b square plus b four and plus a b square minus minus a square. And the third one is minus a. So it's minus a cube plus a b square plus a square. And we still have these three members plus, plus four a square b square minus four a b square plus b square. Okay. That's my overall expression. Well, it's a little long, but, well, what can you do? Just to make things simpler, we can do the following. Look at this. a to the fourth plus four a square b square plus b to the fourth. This obviously is a square plus b square square. But a square plus b square is one. So the combination of this, this and this gives me one because it's a one square, which is one. Again, a square plus b square square is a to the fourth plus four a square b square plus, well, actually, I'm sorry, that's not exactly true. That should include this as well. Minus two a square b square and minus, because minus a b square a square b square and minus a square b square and plus four a a square b square. All together is two plus two a square b square. And that's what gives me one. Because a square plus b square square will be a to the fourth plus two a square b square, which is this, this and this. Minus one minus one plus four. So it's two a square b square and b to the fourth. So that's one. Okay. I told you it's a little tedious and not very interesting. Anyway, what else do we have here? Okay. Okay. What can we do here? All right. So minus a cube remains. The last thing we can do about this. Now a b square. Oh, there's another minus a cube. Okay. So we will put two a cube here, two a cube. So we covered this guy and this guy. Okay. Great. Now, plus a b square plus a b square and minus four a b square. So it's minus two a b square. So we covered this, we covered this, and we covered this. What's remaining? a square and b square, which are together again equal to one. So we have another one. So instead of that, I will put two here. That's what we have, right? Now this can be simplified even further. So this is two minus two a and in parentheses a square plus b square. Right? If I a, bring it outside of the parentheses minus a cube and minus two a b square, which is equal to, again, as we know, a square plus b square is one. So it's two minus two a. So you see, such a very big and complex, complex in terms of complexity, not complex numbers. Such a big expression actually simplified to relatively simple form. Okay. So that's my module of the difference between square of the complex number and complex number itself, original. Okay. So let's just put it somewhere here. I don't need this. I can just write the answer. Two minus two a. Very nice, very simple, etc. Okay, great. Now b, what's the module of the difference between our original complex number and the point on the circle which lies on the x-axis. Remember? So this is a and this is b. So this is the length which I am trying to establish right now, which is modular of the number b, which is square of this plus square of this, a square minus two a plus one plus b square equals, again, we all know that a square plus b square is one and another one, so it's two minus two a. As you see, they are equal, which basically proves the point. Whenever you square this number, which is represented by a point on the unit circle, you just have to double the angle from this number to the x-axis. So this is two alpha. That proves the point. Thank you very much.