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Question Number 27382 by abdo imad last updated on 05/Jan/18

resolve inside C (((z−i)/(z+i)))^n +(((z+i)/(z−i)))^n = 2cosθ and0 <θ<π .n integer.

$${resolve}\:{inside}\:{C}\:\:\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} +\left(\frac{{z}+{i}}{{z}−{i}}\right)^{{n}} =\:\mathrm{2}{cos}\theta\:{and}\mathrm{0}\:<\theta<\pi\:.{n}\:{integer}. \\ $$

Answered by sma3l2996 last updated on 05/Jan/18

(((z−i)/(z+i)))^n +(((z−i)/(z+i)))^(−n) −2cosθ=0 (((z−i)/(z+i)))^(2n) +1−2(((z−i)/(z+i)))^n cosθ=0 ((((z−i)/(z+i)))^n )^2 −2(((z−i)/(z+i)))^n cosθ+(cosθ)^2 +sin^2 θ=0 ((((z−i)/(z+i)))^n −cosθ)^2 +sin^2 θ=0 (((z−i)/(z+i)))^n −cosθ=(√((−sin^2 θ)))=+_− isinθ=isinθ (because 0≤θ≤π) so (((z−i)/(z+i)))^n =cosθ+isinθ=e^(iθ) ((z−i)/(z+i))=e^(i(θ/n)) ⇔z−i−(z+i)e^(i(θ/n)) =0⇔z(1−e^(i(θ/n)) )=i(1+e^(i(θ/n)) ) z=i((1+e^(i(θ/n)) )/(1−e^(i(θ/n)) ))

$$\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} +\left(\frac{{z}−{i}}{{z}+{i}}\right)^{−{n}} −\mathrm{2}{cos}\theta=\mathrm{0} \\ $$ $$\left(\frac{{z}−{i}}{{z}+{i}}\right)^{\mathrm{2}{n}} +\mathrm{1}−\mathrm{2}\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} {cos}\theta=\mathrm{0} \\ $$ $$\left(\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} \right)^{\mathrm{2}} −\mathrm{2}\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} {cos}\theta+\left({cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$ $$\left(\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} −{cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$ $$\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} −{cos}\theta=\sqrt{\left(−{sin}^{\mathrm{2}} \theta\right)}=\underset{−} {+}{isin}\theta={isin}\theta\:\left({because}\:\mathrm{0}\leqslant\theta\leqslant\pi\right) \\ $$ $${so}\:\:\left(\frac{{z}−{i}}{{z}+{i}}\right)^{{n}} ={cos}\theta+{isin}\theta={e}^{{i}\theta} \\ $$ $$\frac{{z}−{i}}{{z}+{i}}={e}^{{i}\frac{\theta}{{n}}} \Leftrightarrow{z}−{i}−\left({z}+{i}\right){e}^{{i}\frac{\theta}{{n}}} =\mathrm{0}\Leftrightarrow{z}\left(\mathrm{1}−{e}^{{i}\frac{\theta}{{n}}} \right)={i}\left(\mathrm{1}+{e}^{{i}\frac{\theta}{{n}}} \right) \\ $$ $${z}={i}\frac{\mathrm{1}+{e}^{{i}\frac{\theta}{{n}}} }{\mathrm{1}−{e}^{{i}\frac{\theta}{{n}}} } \\ $$

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