# Help with De Moivre's Theorem: Complex Numbers

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.

First prove that $z \ne 0$ and then divide both sides of the equation by $z^n$. Then note that $\frac{z+i}{z} = 1 + \frac{i}{z}$.
• @SheowBoon: Show how you get $\frac{i}{z} = 0$. That's not possible. Jun 20 '16 at 14:13
• @SheowBoon: Ah I know your mistake. $(?)^n = 1$ does not imply $? = 1$. Jun 20 '16 at 14:14
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