How to solve this complex number equation: $(z+1)^n-(z-1)^n=0$

Question

I am having trouble with this problem, for some time now.

$$(z+1)^n-(z-1)^n=0$$

Solution is: $$z_k=-i\;ctg\!\left(\frac{k \pi}{n}\right); \: k=1,2,\dots,n-1$$

Can someone explain to me, how to get to this solution?

Answer 1

$$1=e^{2ik\pi}$$ $$\left(\frac{z+1}{z-1}\right)^n=t^n=1 \implies t=e^{2ik\pi/n}, k=0,1,2,..n-1$$ $$\implies z= \frac{t+1}{t-1}=\frac{e^{2ik\pi/n}+1}{e^{2ik\pi/n}-1}=\frac{e^{i k\pi/n}+e^{-ik\pi/n}}{e^{i k\pi/n}-e^{-ik\pi/n}}=-i \cot (k \pi/n)= z_k$$