# Nth roots of a Complex Number

I am currently progressing through the study of Complex Numbers! I'm finding it enjoyable but there is something I just don't understand. Take this example: Calculate $$z^4 = 1$$ Now, I'm going to describe my thought process in the solving of this problem, if I have a misconception about anything please pull me up! Since $\mathbb{R} \subset \mathbb{C}$, the equation can be represented as $$z^4 = 1+0i$$ expressing this as complex exponentials gives us $$(re^{i\theta})^4 = 1e^{i0}$$ $$r^4e^{i4\theta} = 1e^{i0}$$ $$r^4 = 1 , 4\theta = 0 + 2\pi k$$ $$\therefore r = 1, \theta = \frac{2\pi k}{4}$$

Q1: Here is the first idea, I don't fully grasp. So, I comprehend the idea that in order for two complex numbers to be equal, their size must be equal, but I'm not sure about the second part. Is it the angle they make with the positive x-axis must be the same?

Q2: The second thing I don't quite understand is this 'k' value. I understand that it can be thought of as a parameter. But I've always understood 'k' to be the amount of revolutions around the unit circle. So, $k = 0$ would mean that we are dealing with an angle that is $<2\pi$. $k=1$ meaning we are dealing with an angle that is $<4\pi$ and so on and so forth. But I find this idea really weird when dealing with this topic. I know that the solutions to this equation would be $1,-1,i,-i$ and obtaining these values would done so by letting $k=0,1,2,3$. But the angle between each of these complex numbers is $\frac{\pi}{2}$ when they should be $2\pi k$ if we let $k=0,1,2,3$

I tried to make this as clear as possible, but I find it very hard to do so because I have to describe what I'm actually thinking in my head. So, I really hope this made sense!

• You don't need to express z as an exponential. Note that $1 = \cos(2\pi k) + i\sin(2 \pi k) = e^{2 \pi ki}$. The answer follows. Doing the question this way might help with some of the issues. – Kaynex Jan 14 '17 at 19:14
• The angle between the complex numbers is $2 \pi k$ ONLY BEFORE you divide by 4. – Kaynex Jan 14 '17 at 20:01
• For your Q1, maybe you can try proving by contradiction, to get a feel of it: Let $z_1,z_2 \in \Bbb C, \lVert z_1 \rVert = \lVert z_2 \rVert, \arg z_1 \neq \arg z_2$. Suppose that $z_1 = z_2$... – f1garo Jan 14 '17 at 20:08

Probably, thinking in terms of general solutions of trigonometric equations helps.

$$z^4 = r^4 e^{i4\theta} = r^4 (\cos 4\theta + i\sin 4\theta).$$

But $z^4 = 1 = 1 + 0i$. Now, $|z^4| = |1 + 0i| \implies r^4 = 1 \implies r = 1$. Then, we get,

$$\cos 4\theta = 1.$$

This happens only at even multiples of $π$ ($\cos x = -1$ for odd multiples). Hence,

$$4\theta = 2kπ \implies \theta = \frac{kπ}{2}.$$

where $k \in \mathbb{Z}$. In general, if $z^n = 1$, then

$$\cos n\theta = 1 \implies \theta = \frac{2kπ}{n}$$

where $k \in \mathbb{Z}$.