If \(\displaystyle n\in\mathbb{N}, n\geq2\) and \(\displaystyle S=\big \{z\in\mathbb{C}| (z+i)^{n}=(z-i)^{n} \big \}\) then \(\displaystyle S=?\) The right answer is
\(\displaystyle
S=\left \{\operatorname{ctg}\frac{k\pi}{n} |1\leq k\leq n-1;k\in\mathbb{N}\right \}
\)
I started like this \(\displaystyle \frac{z-i}{z+i}=\cos(\frac{2k\pi}{n})+i\sin(\frac{2k\pi}{n})\). How to continue? Some ideas?
\(\displaystyle
S=\left \{\operatorname{ctg}\frac{k\pi}{n} |1\leq k\leq n-1;k\in\mathbb{N}\right \}
\)
I started like this \(\displaystyle \frac{z-i}{z+i}=\cos(\frac{2k\pi}{n})+i\sin(\frac{2k\pi}{n})\). How to continue? Some ideas?