$$\overline{z}=z^n$$ where $n\in \mathbb{N}$

So I have started with moving the polar representation as the expression is in the n-th power


I can not multiply both sides in $r^{(-n)}$ as $n\in \mathbb{N}$ how should I continue?


marked as duplicate by dxiv, Did, Claude Leibovici, kingW3, C. Falcon Mar 23 '17 at 19:55

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For $n=1$, the answer is that $z$ can be any real number.

For $n>1$, observe that $|\overline{z}|=|z|$ so we must have $|z|=|z^n|$ which means that $z=0$ or $|z|=1$ (in $r\cdot cis\theta$ notation, $r=1$). This reduces the problem to a problem of rotations around the unit circle which can be easily solved.

  • $\begingroup$ So overall we have $n$ solutions? $\endgroup$ – gbox Mar 22 '17 at 20:27
  • $\begingroup$ @gbox $n+1$ solutions. The $n$ solutions on the unit circle, and also $0$. $\endgroup$ – Stella Biderman Mar 22 '17 at 20:28
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    $\begingroup$ @gbox $0$ is not on the unit circle because the unit circle is the set of all points with norm $1$. $\endgroup$ – Stella Biderman Mar 22 '17 at 20:32
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    $\begingroup$ @dxiv right! Forgot about incrementing the exponent. $\endgroup$ – Stella Biderman Mar 22 '17 at 20:33
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    $\begingroup$ You have $\overline{z}=z^n$. Taking the norm of both sides gives $|\overline{z}|=|z^n|$. The LHS is known to equal $|z|$ and the RHS is known to equal $|z|^n$. $\endgroup$ – Stella Biderman Mar 22 '17 at 20:36

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