Complex Unity and Solving n- complex equations , SG -Level

Exercise:

Solve in $\mathbb{C},$ ${{\left( z+i \right)}^{n}}={{\left( z-i \right)}^{n}}$ where $n\in {{\mathbb{N}}^{*}}$

Solution: if ${{a}^{n}}={{b}^{n}}\Leftrightarrow {{\left( \frac{a}{b} \right)}^{n}}=1$

So we have ${{\left( z+i \right)}^{n}}={{\left( z-i \right)}^{n}}\Leftrightarrow {{\left( \frac{z+i}{z-i} \right)}^{n}}=1$

Note that this equation is defined when $z\ne \pm i$

Put  $Z=\frac{z+i}{z-i}$ thus ${{Z}^{n}}=1$

Let $Z=r{{e}^{i\theta }}\Leftrightarrow {{Z}^{n}}={{r}^{n}}{{e}^{in\theta }}={{e}^{i0}}$

So ${{r}^{n}}=1\,\,\And \,\,n\theta =0+2k\pi \Leftrightarrow r=1\,\,\,\And \,\,\theta =\frac{2k\pi }{n}\,\,\,\,\,\,,\,\,\,0\le k\le n-1$

But $Z=\frac{z+i}{z-i}\Leftrightarrow zZ-iZ=z+i\Leftrightarrow z\left( Z-1 \right)=iZ+i$

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

Note that ${{\cos }^{2}}\theta =\frac{1+\cos 2\theta }{2}\,\,\,\,\And \,\,\,\,{{\sin }^{2}}\theta =\frac{1-\cos 2\theta }{2}$

So $1+\cos \theta =2{{\cos }^{2}}\left( \frac{\theta }{2} \right)\,\,\,\And \,\,\,1-\cos \theta =2{{\sin }^{2}}\left( \frac{\theta }{2} \right)$

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

        $\,\,\,\,\,=i\frac{\cos \left( \frac{\theta }{2} \right)\left[ \cos \left( \frac{\theta }{2} \right)+i\sin \left( \frac{\theta }{2} \right) \right]}{i\sin \left( \frac{\theta }{2} \right)\left[ \cos \left( \frac{\theta }{2} \right)+i\sin \left( \frac{\theta }{2} \right) \right]}$

So $z=\cot \left( \frac{\theta }{2} \right)=\cot \left( \frac{k\pi }{n} \right)\,\,\,\,\,,\,\,\,\,\,\,\,1\le k\le n-1$ ,$k\in \mathbb{Z}$