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I have a homework problem which goes:

Given $z^n=(z+i)^n$, using de Moivre's Theorem, show that

$z=\frac{i}{e^\frac{i2k\pi}{n}-1}$

What steps should I take in tackling this question? It's a 2 mark question and I can't seem to find an appropriate way to solve it quickly.

Thanks in advance!

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First prove that $z \ne 0$ and then divide both sides of the equation by $z^n$. Then note that $\frac{z+i}{z} = 1 + \frac{i}{z}$.

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  • $\begingroup$ Hi, sorry but what do I do to prove the value of z after that? Thanks $\endgroup$
    – Sheow Boon
    Jun 20 '16 at 13:57
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    $\begingroup$ @SheowBoon: Well, first tell me what you get after that. Math SE is not here to do your homework for you, but to guide you to solve problems by yourself as far as possible. $\endgroup$
    – user21820
    Jun 20 '16 at 14:03
  • $\begingroup$ What I get is i/z=0, so I could then derive that z=i/0, but I'm not sure why z would equal to infinity, if it still is a complex number. $\endgroup$
    – Sheow Boon
    Jun 20 '16 at 14:08
  • $\begingroup$ @SheowBoon: Show how you get $\frac{i}{z} = 0$. That's not possible. $\endgroup$
    – user21820
    Jun 20 '16 at 14:13
  • $\begingroup$ @SheowBoon: Ah I know your mistake. $(?)^n = 1$ does not imply $? = 1$. $\endgroup$
    – user21820
    Jun 20 '16 at 14:14
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Take $z^n$ common from Rhs cancelling $z^n$ as $|z|\neq 0$ we get $(1+\frac{i}{z})^n=1$ now let $(1+..)=l$ so we get $l^n=1$ thus $l=e^{\frac{i2k\pi}{n}}$ so $z=..$ we get the desired result

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