Complex Numbers Problems

Question

I have a problem that I need help with - Find the number of complex numbers $z$ such that $z^{2018} =\overline{z}$

So far, I've converted each side to complex exponential form ($re^{n*i*\theta}$), but I'm not sure how to continue...

Thanks! Help is appreciated!

Answer 1

You could say that $$z = re^{i \theta}$$ So the equation becomes $$r^n e^{in\theta} = re^{-i\theta}$$ where $n = 2018$. This means that $r = \lbrace 0,1 \rbrace$ and $$n\theta = -\theta + 2k\pi \qquad k \in \lbrace 0\ldots n \rbrace$$ or $$(n+1)\theta = 2k\pi \qquad k \in \lbrace 0\ldots n \rbrace $$ which means that $$\theta = \frac{2k\pi}{n+1} \qquad k \in \lbrace 0\ldots n \rbrace$$ which gives you $$n-0+1 = n+1 = 2018+1 = 2019$$ So in total you have $2019 + 1 = 2020$ solutions.

Answer 2

One solution to $z^n=\bar{z}$ is $z=0$ (where $n$ is any positive integer). Suppose $z\ne0$; then $$ |z^{n}|=|\bar{z}| $$ implies $|z|=1$ (why?).

Therefore $\bar{z}=z^{-1}$ and the equation becomes $$ z^{n+1}=1 $$

How many $(n+1)$-th roots of unity are there?