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Solve the following equation:

$\bar{z}=z^{n-1}$

Where $\bar{z}$ is the complex conjugate of z, and n is a natural number such that $n\neq 2$.

I have tried to write z in rectangular form and polar form. I have tried to play with De Moivre's formula.

But I still do not see where to proceed from here.

Could you please point me to the right direction?

Thanks.

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  • $\begingroup$ Multiply both sides by $z$, and taking absolute values on both sides gives $|z|^2=|z|^n$, so either $|z|=1$ or $n=2$, which is contrary to your assumption. $\endgroup$ – Adam Hughes Sep 9 '14 at 3:39
  • $\begingroup$ Does this imply that the only complex number satisfying this equation is any complex number such that its magnitude is 1? $\endgroup$ – user101998 Sep 9 '14 at 3:52
  • $\begingroup$ Not all of them work either, this is a necessary but not sufficient condition. You only asked how to get started, so this is a push towards the full solution. $\endgroup$ – Adam Hughes Sep 9 '14 at 3:53
  • $\begingroup$ Hint: $z^n = z^{n-1}z = \bar{z}z = |z|^2$. $\endgroup$ – achille hui Sep 9 '14 at 3:54
  • $\begingroup$ Okay, let's see what I can do. Thanks! $\endgroup$ – user101998 Sep 9 '14 at 3:59
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First, multiply both sides by $z$ to find $$ |z|^2 = z^n $$ So, in particular, $z^n$ needs to be a non-negative real number. So, we can say that the argument of $z$ must be a multiple of $2 \pi/n$.

Now, how about the magnitude? Taking the magnitude on both sides of the original equation gives us $$ |z| = |z|^{n-1} \implies|z|^{n-1} - |z| = 0 \implies\\ |z|\cdot (|z|^{n-2} - 1) = 0 $$ So, what are the possibilities for $|z|$?

Together, the two pieces of information should be enough to figure out all possibilities for $z$.

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Use polar coordinates, $z = re^{i\theta}$: $$re^{-i\theta} = r^{n-1}e^{i(n-1)\theta},$$ $$r = r^{n-1}\implies\cdots,$$ $$\cdots$$

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