<s>[INST] Finding complex roots of $(z-i)^n-(z+i)^n =0$ I have a problem. First part of it was to find $z$ such that $$\left|\frac{z-i}{z+i}\right| = 1$$ and I've quickly figured out, that $z \in \Bbb R$. 
How ever, now I have to (using previous part of the problem) find all roots of the following equation: $$(z-i)^n - (z+i)^n = 0$$ 
I have no idea how to do this. I tried adding the second part to the right side, then dividing by it, but with no success. 
 [/INST] $$(z-1)^n-(z+1)^n=0\ (n\ \epsilon\ \mathbb{Z}^+)$$
$$\Rightarrow (z-1)^n=(z+1)^n \Rightarrow \left(\frac{z-1}{z+1}\right)^n=1=1\angle 2\pi k\ (k=0,1,2,...n-1)$$
$$\Rightarrow \frac{z-1}{z+1}=1\angle \frac{2\pi k}{n}\Rightarrow 1 -\frac{2}{z+1}=1\angle \frac{2\pi k}{n}$$
$$\frac{2\pi k}{n}=\theta \Rightarrow 1 -\frac{2}{z+1}=1\angle \theta \Rightarrow z=\frac{1+1\angle \theta}{1-1\angle \theta}$$
$$\Rightarrow z = \frac{1+i\sin \theta + \cos \theta}{1-i\sin \theta - \cos \theta} = \frac{(1+i\sin \theta + \cos \theta)^2}{1-(i\sin \theta + \cos \theta)^2}$$
$$\Rightarrow z= \frac{2\cos^2 \theta + 2i\sin \theta \cos \theta +2i\sin \theta + 2\cos \theta}{2\sin^2 \theta -2i\sin \theta \cos \theta}$$
$$\Rightarrow z=\frac{2(i\sin \theta + \cos \theta)(\cos \theta + 1)}{-2i\sin \theta (i\sin \theta + \cos \theta)}=\frac{i(\cos \theta +1)}{\sin \theta}$$
$$\Rightarrow z = \csc \theta + i\cot \theta$$
$$\Rightarrow z = \csc \frac{2\pi k}{n} + i\cot \frac{2\pi k}{n}\ (k=0,1,2,...n-1)$$
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