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I need to find all solutions of such complex number equation:

$z^{13}=\overline{z}$

We assume that $r = |z|$ and we use Euler's formula $z=|z|e^{i\phi}=re^{i\phi}$.

Then, multiplying both sides by $z$, we have

$z^{14}=|z|^2$

$z^{14}-|z|^2 = 0$

$r^{14}e^{14i\phi}-r^2 = 0$

$r^2(r^{12}e^{14i\phi}-1) = 0 $

Our solutions will be: $r^2 = 0$ and $r^{12}e^{14i\phi}=1$.

While from first solution I can see that $z=0$, then I'm not sure how to correctly and legally solve the second solution. Can I cancel the $r^{12}$?

WLOG
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stil
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    Since $r^{12} = \lvert r^{12}e^{14\phi}\rvert = \lvert 1\rvert = 1$ and $r \geqslant 0$, it follows that $r = 1$. So yes. – Daniel Fischer Feb 10 '14 at 12:47

2 Answers2

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Suppose $z \neq 0$, $ z = re^{i \phi}$

$ z^{13}=\overline{z} \Rightarrow z^{14}=|z|^{2} $

So $Arg(z^{14}) = 2k\pi \ \ \ , k \in \mathbb{Z} \Rightarrow Arg(z) = \frac{k \pi}{7} , k \in \mathbb{Z} $

And $r^{14} = r^{2} \Rightarrow r^{2}(r^{12} -1) = 0 \Rightarrow r = 1$

So $$ z = e^{i\phi} \ \ \ , \phi = \frac{k \pi}{7} \ \ \ , k \in \mathbb{Z}$$ and $z = 0$

WLOG
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Obviously, $z=0$ is a solution.

Assume now $z\neq 0$. Taking the norm we have: $$|z|^{13}=|z^{13}|=|\bar{z}|=|z|$$ Thus $|z|=1$. In particular, this implies that $\bar{z}=z^{-1}$, and thus the original equation can be written as: $$z^{14}=1$$ The solutions of this equation are the $14$ roots of unity $e^{2\pi i\frac{k}{14}}$, where $k$ ranges from $0$ to $13$.

Daniel Robert-Nicoud
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