Taking and visualizing powers of a complex number | Precalculus | Khan Academy

We're told to consider the complex number ( z ) is equal to negative one plus ( i ) times the square root of three. Find ( z ) to the fourth in polar and rectangular form. So pause this video and see if you can figure that out.

All right, now let's work through this together. So first, let's just think about what the modulus of ( z ) is. We know that the modulus is going to be equal to the square root of the real part squared plus the imaginary part squared. So it is going to be ( (-1)^2 + (\sqrt{3})^2 ), which is going to be equal to ( 1 + 3 ), so the principal root of 4, which is equal to 2.

Now, the next interesting question is what is the argument of ( z ). The reason why I'm even going through this is once we put it into polar form, it's going to be a lot easier to both visualize what it means to take the various exponents of it and then we can convert back into rectangular form.

So let me draw another complex plane here: imaginary axis, that is my real axis. If I were to plot ( z ), it would look something like this. We have negative 1 in the real direction, so that might be negative 1 there, and we have ( \sqrt{3} ) in the imaginary direction. So our point ( z ) is right over here.

We know the distance from the origin, the modulus, and we know that this distance right over here is 2. We know that this distance right over here is ( \sqrt{3} ) and we know that this distance right over here is 1. You might immediately recognize this as a 30-60-90 triangle because, in a 30-60-90 triangle, the short side is half of the hypotenuse, and the long side is the square root of 3 times the short side.

So we know that this is a 60 degree angle. We know that this is a 30 degree angle. The reason why that helps us, it's hard to see that 30 degree. The reason why that helps us is that this is 60 degrees. We know that the argument here must be 120 degrees. So the ( \text{arg} ) of ( z ), the argument of ( z ), is 120 degrees.

Just like that, we can now think about ( z ) in polar form. So let me write it right over here: we can write that ( z ) is equal to its modulus 2 times the cosine of 120 degrees plus ( i ) times the sine of 120 degrees. We could also visualize ( z ) now over here. So its modulus is 2, so that's halfway to 4, and its argument is 120 degrees, so it would put us right over here. This is where ( z ) is.

Now what would ( z^2 ) be? Well, when you multiply complex numbers and you've represented them in polar form, we know that you would multiply the moduli. So it would then be ( 2^2 ), so it'd be 4 right over here, and then you would add the arguments. So you would essentially rotate ( z ) by another 120 degrees because you're multiplying it by ( z ). So it's going to be cosine of 240 degrees plus ( i ) sine of 240 degrees. Once again, ( 2 \times 2 ) is equal to 4, and ( 120 ) degrees plus another ( 120 ) degrees is ( 240 ) degrees.

Now, where would ( z^2 ) sit? Well, its argument is ( 240 ) degrees and its modulus is 4, so now it is twice as far from the origin. Let's think about what... I'll do this in a new color. What ( z ) to the third power is going to be equal to? Well, that's going to be ( z^2 ) times ( z ) again. So we're going to multiply 2 times this modulus, so that's going to be equal to 8, and then we're going to rotate ( z^2 ) by 120 degrees: cosine of 360 degrees plus ( i ) sine of 360 degrees.

So that's going to put us at 8 for our modulus, and ( 360 ) degrees is the same thing as ( 0 ) degrees, so we are right over here. So this is ( z ) to the third power, and I think you know where this is going. What is ( z ) to the fourth power going to be? Let me move my screen down a little bit so I have a little more real estate.

( z^4 ), well I'm just going to take this modulus here since I'm going to multiply ( z^3 ) times ( z ). I'm going to multiply that modulus times 2 to get to 16, and then I'm going to add another 120 degrees. Well, I could write ( \cos(480) ) degrees or ( 360 ) degrees, the same thing as ( 0 ) degrees.

So this I could say is ( 0 ) degrees. This is ( 0 ) degrees, so if I add ( 120 ) to that, I get ( \cos(120) ) degrees plus ( i \sin(120) ) degrees. So my argument is back to being at ( 120 ) degrees but now my modulus is 16. So there's 4, 8, 12, 16. This outer circle right over here, I am right over there with ( z ) to the fourth.

We're almost done. We've just represented ( z ) to the fourth in polar form. Now we just have to think about it in rectangular form. Now, lucky for us, we already know what ( \cos(120) ) degrees is and ( \sin(120) ) degrees is. We can construct, if we want, another 30-60-90 triangle right over here.

So the hypotenuse here has length 16, the short side is going to be half of that, so it has length 8, and then the long side is going to be ( \sqrt{3} ) times the short side, so it's going to be ( 8\sqrt{3} ). If we wanted to write ( z ) to the fourth in a rectangular form, it would be the real part: ( -8 + i \times 8\sqrt{3} ).

And we're done.