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!
Best Answer
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.