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I have encountered such a question that I am not sure how to solve it:

$z= a+bi$, where a and b are real numbers, solve the equation $\bar{z}^5 = z$ ($\bar{z}$ is the conjugate of $z$)

  • The question also gives two hints: $cos(\pi/3)=\frac{1}{2}$ and $sin(\pi/3) = \frac{\sqrt{3}}{2}$

However, I am not sure how to use these two hints?

I managed to find the solution in a reverse way: I found that if $z =\frac{1}{2}+\frac{\sqrt{3}}{2}i $ would satisfy the equation of this question.

But I was not able to find this solution from the equation $\bar{z}^5 = z$

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    $\begingroup$ Why not use polar form instead? $\endgroup$ Commented Dec 19, 2019 at 0:39

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If you multiply both sides by $\overline z$ you get $$\overline z^6=|z|^2$$ The right side is real and nonnegative, so the left side must be as well. The magnitude of the left is $|z|^6$ so we know that $|z|=0$ or $1$. $\overline z$ and therefore $z$ must be $0$ or one of the sixth roots of $1$.

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