Find the number of complex numbers $z$ such that $z^{2018} =\overline{z}.$

Question

Find the number of complex numbers $z$ such that $$z^{2018} =\overline{z}.$$

I have the basic idea of plugging in $z$ as $a+bi$ but it seems as if it is a dead end. $(a+bi)^{2018} = a-bi$ doesn't lead me anywhere. I am puzzled at how to start.

Answer 1

Hint: $z=0$ is a solution. In order to find the rest of the solutions you may assume that $z\not=0$ and so you can multiply both sides by $z$ this implies that

$z^{2019} = z\overline{z}=|z|^2$

Now if $|z|=r$ then $|z^{2019}|=|z|^{2019}$ and so the equation implies that $|z|=1$ (because $z\not=0$).

Therefore you only have to solve the equation $z^{2019}=1$. The solutions are of course all $2019$ roots of unity in the circle.

Answer 2

If $z^{2018}=\overline z$,$$|z|^{2018}=\bigl|\overline z\bigr|=|z|$$and therefore $|z|=0$ or $|z|=1$. But $|z|=1\implies\overline z=\frac1z$. So, the solutions are $0$ and the complex numbers $z$ such that $z^{2019}=1$. And the solutions of this last equation are the numbers of the form$$\exp\left(\frac{2k\pi i}{2019}\right)\text{, with }k\in\{0,1,\ldots,2018\}.$$

Answer 3

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 $r = 1$ 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 $$n-0+1 = n+1 = 2018+1 = 2019$$ solutions

Answer 4

I'll treat this in some generality, looking for solutions to

$z^n = \bar z, \; 2 \le n \in \Bbb N; \tag 1$

First of all observe that $z = 0$ solves (1); so we may refine our search to non-vanishing $z$; then from (1)

$\vert z \vert^n = \vert z^n \vert = \vert \bar z \vert = \vert z \vert; \tag 2$

or

$\vert z \vert^{n - 1} = 1 \Longrightarrow \vert z \vert = 1; \tag 2$

therefore, we may write

$z = e^{i\theta}, \; \theta \in [0, 2\pi); \tag 3$

then

$\bar z = e^{-i\theta} = z^{-1}, \tag 4$

so that (1) yields

$z^n = z^{-1}, \tag 5$

or

$z^{n + 1} = 1; \tag 6$

then

$(e^{i\theta})^{n + 1} = 1 \tag 7$

or

$e^{(n + 1)i \theta} = 1; \tag 8$

we may thus take

$(n + 1) \theta = 2\pi, \tag 9$

whence

$\theta = \dfrac{2\pi}{n + 1}; \tag{10}$

this is the smallest non-zero $\theta \in [0, 2\pi)$ such (1) binds; of course, we in fact have

$(e^{2k\pi i/ (n + 1})^{n + 1} = e^{2k\pi i} = (e^{2\pi i})^k = 1^k = 1, \; 1 \le k \le n + 1, \tag{11}$

so there are in fact $n + 1$ solutions with $\vert z \vert = 1$; and if we add to this collection the $0$ solution, we see there are $n + 2$ solutions to (1) in toto. We further observe that since $e^{2k\pi i / (n + 1)} = (e^{2\pi i / (n + 1)})^k$, every non-vanishing solution is an integral power of $e^{2\pi i /(n + 1)}$.

We have restricted ourselves so far to $2 \le n \in \Bbb N$; when $n = 1$, (1) becomes

$z = \bar z, \tag{12}$

and any $z \in \Bbb R$ is a solution, so they are uncountable in number in this case.

When $n = 2018$, there are 2020 solutions:

$z = 0, z = e^{2k \pi / (n + 1)}, \; 1 \le k \le 2019. \tag{13}$