Complex exercise about Z^n=-1 (unity)

Exercise:

Solve in$\mathbb{C}$,${{\left( z+i \right)}^{n}}+{{\left( z-i \right)}^{n}}=0$  for $z\ne i$

Solution: we have ${{\left( z+i \right)}^{n}}+{{\left( z-i \right)}^{n}}=0\Leftrightarrow {{\left( z+i \right)}^{n}}=-{{\left( z-i \right)}^{n}}$

The above equation is defined when $z\ne i$

$\Leftrightarrow {{\left( \frac{z+i}{z-i} \right)}^{n}}=-1\Leftrightarrow {{Z}^{n}}=-1\,\,\,\,\,,\,\,\,\,Z=\frac{z+i}{z-i}$

Let $Z=r{{e}^{i\theta }}\Leftrightarrow {{Z}^{n}}={{r}^{n}}{{e}^{in\theta }}={{e}^{i\pi }}$

Thus ${{r}^{n}}=1\,\,\,\,\And \,\,\,n\theta =\pi +2k\pi \Leftrightarrow r=1\,\,\,\And \,\,\theta =\frac{\pi +2k\pi }{n}$ , $1\le k\le n-1\,,\,\,k\in \mathbb{Z}$

We have $Z=\frac{z+i}{z-i}\Leftrightarrow zZ-Zi=z+i\Leftrightarrow zZ-z=Zi+i\Leftrightarrow z=\frac{i\left( Z+1 \right)}{Z-1}$

$\Rightarrow z=i\frac{{{e}^{i\theta }}+1}{{{e}^{i\theta }}-1}=i\frac{\cos \theta +i\sin \theta +1}{\cos \theta +i\sin \theta -1}=i\frac{2{{\cos }^{2}}\left( \frac{\theta }{2} \right)+i2\sin \left( \frac{\theta }{2} \right)\cos \left( \frac{\theta }{2} \right)}{-2{{\sin }^{2}}\left( \frac{\theta }{2} \right)+i2\sin \left( \frac{\theta }{2} \right)\cos \left( \frac{\theta }{2} \right)}$

$\Rightarrow z=i\frac{2\cos \left( \frac{\theta }{2} \right)\left[ \cos \left( \frac{\theta }{2} \right)+i\sin \left( \frac{\theta }{2} \right) \right]}{-2\sin \left( \frac{\theta }{2} \right)\left[ -{{i}^{2}}\sin \left( \frac{\theta }{2} \right)+i\cos \left( \frac{\theta }{2} \right) \right]}=i\cot \left( \frac{\theta }{2} \right)\frac{{{e}^{i\left( \theta /2 \right)}}}{i\left( \cos \left( \frac{\theta }{2} \right)+i\sin \left( \frac{\theta }{2} \right) \right)}$

$\Rightarrow z=\cot \left( \frac{\theta }{2} \right)\frac{{{e}^{i\left( \theta /2 \right)}}}{{{e}^{i\left( \theta /2 \right)}}}=\cot \left( \frac{\theta }{2} \right)=\cot \left( \frac{\frac{\pi +2k\pi }{n}}{2} \right)=\cot \left( \frac{\pi +2k\pi }{2n} \right)$ , $1\le k\le n-1\,\,,\,k\in \mathbb{Z}$