Complex $n$th Roots

Hello everyone. In this video, we're going to talk about complex $n$th roots. In particular, complex third roots, or complex cube roots, because that's the problem that I picked.

Find the cube roots of $1-2i$. (You may leave part of your answer in terms of $\arctan$).

The answer to this question is going to be ugly and use a lot of notation. It's just the way it's going to be. I didn't want to try to fudge the numbers so that they were correct. I just decided hey let's just leave it and let's just talk about the ideas because again the key ideas are what are really important in these examples, as opposed to making the numbers look perfect.

Solution

So what are we going to do in this question? So we're trying to find for the cube roots of $1-2i$, so we're trying to solve for $z^3=1-2i$, and when we're solving for $z^3=1-2i$, it's easier to deal with polar coordinates than it is to deal with standard form.

So we're going to let $z=re^{i\theta}$. Then what do we have?

$$r^3e^{i3\theta}=1-2i$$

From here we have options. What I'm going to do is I'm going to determine $r$ first. $r$ is always easy to determine because you can determine the lengths of both sides. So by PM,

$$|r^3e^{i3\theta}|=|r^3||e^{i3\theta}=|r|^3=|1-2i|=\sqrt{(1)^2+(-2)^2}=\sqrt{5}$$

So hence $r=\sqrt[6]{5}$. Remember, $|e^{i3\theta}|=\cos^2(3\theta)+\sin^2(3\theta)=1$. So that's something you should double-check for yourselves, but taking lengths here is actually going to simplify this a lot.

So remember in this notation, $r=|z|$, which has to be a non-negative number, so $r$ is non-negative and it's real so the length is just going to be $+\sqrt[3]{\sqrt{5}}=\sqrt[6]{5}$.

Now we want to solve in terms of $3\theta$. So next we substitute this back into the first displayed equation to get that

$$e^{i3\theta}=r^{-1}=(1-2i)=e^{i\alpha}$$

where $\alpha=\arctan((-2r^{-1})/(r^{-1}))=\arctan(-2)$.

So again, I want to solve for $3\theta$. If you pick your way that you like to solve this, I'm going to convert this to polar coordinates.

Now something to keep in mind, is this the actual correct one? Do we want $\arctan(-2)$ or do we want $\arctan(-2+\pi)$? And again we can answer this, $\arctan(-2)$, I don't know the exact number of this, but that doesn't matter. I know that $\arctan(-2)$ is going to be a negative radian value. So that's going to put us in the fourth quadrant, and $1-2i$ lives in the fourth quadrant, so life is okay. So I don't need to add $\pi$ here, I can just take $\alpha=\arctan(-2)$.

So then what does this give? So if we want to solve for $\theta$, so hence,

$$3\theta=\alpha+2\pi k$$

for any integer $k$.

Again, up to these values, $e^{i2\pi}=1$. So if I take any multiple here, it's going to give me the same value for $e^{i\alpha}$. I mean these two angles must be equal up some multiples of $2\pi$, right, because I can keep rotating around the complex plane, it's going to give me the same angle.

So doing this, and then just solving for $\theta$,

$$\theta=\alpha/3+2\pi k/3$$

So these are our possible solutions. Again, as in class, right, we've already seen that two of these angles are equal if the $k$ values are congruent $\bmod 3$. We don't want to keep doing that. We only want a unique set because they correspond to complex numbers and for complex numbers, eventually you just start to cycle on to yourself. And when does that happen? Every $2\pi$ multiple, right? So when I hit $3$, I hit a $2\pi$ multiple, and so I’m going to go back to when $k=0$, and so on and so forth.

As in class, we have that $k\in\{0,1,2\}$ to give us the $3$ possible solutions.

$$z\in\{\sqrt[6]{5}e^{i(\arctan(-2)/3)},\sqrt[6]{5}e^{i(\arctan(-2)/3+2\pi/3)},\sqrt[6]{5}e^{i(\arctan(-2)/3+4\pi/3)}\}$$

Again if you wanted to, if you're a lot more patient than I'm going to be, you could convert these back to standard form. It's going to take a little bit of effort since we don't know what $\arctan(-2)$ is. If you have a calculator, this is very easy. You can convert it back and forth no problem. As it is though, I’ll leave the answer like this.

Notice that the angles do differ by $2\pi/3$, and that's what's given by the Complex $n$th Roots Theorem, it says once you have a root, all the other ones you just need to rotate by $2\pi/n$ where $n$ was the $z^n$ part, and here $n=3$.

Yeah that's basically it. I mean, that's how you solve something like this. You need to be a little bit careful when you're doing these, but again, the only real part where you could get a little bit tripped up is this $e^{i\alpha}$ part, right, just the same thing as in the previous video about changing to polar coordinates, you might have to worry about adding $\pi$ or not adding $\pi$.

Here again, we know that two of these $\theta$ values are equal if they differ by a multiple of $2\pi$, and so we only need to consider three values for $k$ because after three values of $k$, so after $0$, $1$, and $2$, they start to repeat themselves up to $2\pi$ multiples. You could think of this as modulo $2\pi$, but that would need a lot of explanation that we haven't talked about, but that's the way you can sort of think of it in your head.

And otherwise, I think that's all I want to say. So like cube roots of $1-2i$, not pretty. If you plug these values in they will work, ideally, if I didn't make any typos or anything. But yeah, hopefully this gives you an idea of how to solve these types of problems, how to set them up and how to attack them. So that's great. Thank you very much for your time and good luck.