Question Number 27382 by abdo imad last updated on 05/Jan/18
$${resolve}\:{inside}\:{C}\:\:\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} +\left(\frac{{z}+{i}}{{z}−{i}}\right)^{{n}} =\:\mathrm{2}{cos}\theta\:{and}\mathrm{0}\:<\theta<\pi\:.{n}\:{integer}. \\ $$
Answered by sma3l2996 last updated on 05/Jan/18
$$\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} +\left(\frac{{z}−{i}}{{z}+{i}}\right)^{−{n}} −\mathrm{2}{cos}\theta=\mathrm{0} \\ $$$$\left(\frac{{z}−{i}}{{z}+{i}}\right)^{\mathrm{2}{n}} +\mathrm{1}−\mathrm{2}\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} {cos}\theta=\mathrm{0} \\ $$$$\left(\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} \right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} {cos}\theta+\left({cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\left(\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} −{cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} −{cos}\theta=\sqrt{\left(−{sin}^{\mathrm{2}} \theta\right)}=\underset{−} {+}{isin}\theta={isin}\theta\:\left({because}\:\mathrm{0}\leqslant\theta\leqslant\pi\right) \\ $$$${so}\:\:\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} ={cos}\theta+{isin}\theta={e}^{{i}\theta} \\ $$$$\frac{{z}−{i}}{{z}+{i}}={e}^{{i}\frac{\theta}{{n}}} \Leftrightarrow{z}−{i}−\left({z}+{i}\right){e}^{{i}\frac{\theta}{{n}}} =\mathrm{0}\Leftrightarrow{z}\left(\mathrm{1}−{e}^{{i}\frac{\theta}{{n}}} \right)={i}\left(\mathrm{1}+{e}^{{i}\frac{\theta}{{n}}} \right) \\ $$$${z}={i}\frac{\mathrm{1}+{e}^{{i}\frac{\theta}{{n}}} }{\mathrm{1}−{e}^{{i}\frac{\theta}{{n}}} } \\ $$