$S=\left \{z\in\mathbb{C}| (z+i)^{n}=(z-i)^{n} \right \}$ $S=?$

Question

If $n\in\mathbb{N}, n\geq2$ and $S=\big \{z\in\mathbb{C}| (z+i)^{n}=(z-i)^{n} \big \}$ then $S=?$ The right answer is $$ S=\left \{\operatorname{ctg}\frac{k\pi}{n} |1\leq k\leq n-1;k\in\mathbb{N}\right \} $$

I started like this $\left(\frac{z+i}{z-i}\right)^{n}=1$. How to continue? Some ideas?

Answer 1

Hint: Let be $\omega_k$, $k = 1,\dots,n$ the $n-$th roots of 1. Then, solve each equation $$\frac{z+i}{z-i} = \omega_k.$$

Answer 2

Hint: $z \in S$ implies that $|z+i|=|z-i|$.

Furthermore we have $|z+i|=|z-i| \iff z \in \mathbb R$.

Answer 3

$$\frac{z+i}{z-i} = \frac{x + (y+1)i}{x + (y-1)i} = \frac{x^2 - y^2 + 1}{x^2 + (y-1)^2} + \frac{2xy}{x^2 + (y-1)^2}i = r e^{i\theta}$$ so $$r = \vert \frac{z+i}{z-i} \vert = \sqrt{ \frac{(x^2 - y^2 + 1)^2}{(x^2 + (y-1)^2)^2} + \frac{4x^2y^2}{(x^2 + (y-1)^2)^2} } = 1$$ which is $$ (x^2 - y^2 + 1)^2+4x^2y^2 = (x^2 + (y-1)^2)^2 \tag{1}$$ Another equation could be obtained from the phase, that is $$n\theta = 2k\pi \tag{2}$$ where $\theta = \arctan \frac{2xy}{x^2-y^2 + 1} $

Answer 4

Hint:

You have $\;\dfrac{z-i}{z+i}=\mathrm e^{\tfrac{2ik\pi}n},\enspace k=0,1\dots n-1$.

Solve for $z$, remembering that $\cos x$ and $\sin x$ can also be written as $$\cos x=\frac{\mathrm e^{2ikx}+1}{2\,\mathrm e^{ikx}},\qquad\sin x=\frac{\mathrm e^{2ikx}-1}{2i\,\mathrm e^{ikx}}.$$