Solving $(z+i)^n+(z-i)^n=0$

Question

Given a complex number $z$, how can I solve $$(z+i)^n+(z-i)^n=0$$ for z?

My attempt

I thought that I couldn't possibly convert this into Euler's format. So, I took to Binomial Theorem. $$(z+i)^n+(z-i)^n=0$$ Using binomial theorem, $$2\left[{n \choose 0}z^n+{n \choose 2}z^{n-2}i^2+{n \choose 4}z^{n-4}i^4+...\right]=0$$ $${n \choose 2}z^{n-2}+{n\choose 6}z^{n-6}+...={n \choose 0}z^{n}+{n \choose 4}z^{n-4}+...$$ How can I solve this after this step?

Any hint is appreciated.

Answer 1

Solving:

$$\begin{align}y&=\frac{z+i}{z-i }\\ yz-yi&=z+i \\ (y-1)z&=(y+1)i \\ z&=\frac{iy+i}{y-1} \end{align}$$

Then you need $y^n=-1.$