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

$$resolveinsideC{\left(\frac{z-i}{z+i}\right)}^n+{\left(\frac{z+i}{z-i}\right)}^n=2cos\theta and0<\theta <\pi .ninteger.$$

Answered by sma3l2996 last updated on 05/Jan/18

$${\left(\frac{z-i}{z+i}\right)}^n+{\left(\frac{z-i}{z+i}\right)}^{-n}-2cos\theta =0$$ $${\left(\frac{z-i}{z+i}\right)}^{2n}+1-2{\left(\frac{z-i}{z+i}\right)}^{n}cos\theta =0$$ $${\left({\left(\frac{z-i}{z+i}\right)}^n\right)}^2-2{\left(\frac{z-i}{z+i}\right)}^{n}cos\theta +{\left(cos\theta \right)}^2+{sin}^2\theta =0$$ $${({\left(\frac{z-i}{z+i}\right)}^n-cos\theta )}^2+{sin}^2\theta =0$$ $${\left(\frac{z-i}{z+i}\right)}^n-cos\theta =\sqrt{(-{sin}^2\theta )}=\underset{-}{+}isin\theta =isin\theta (because0⩽\theta ⩽\pi )$$ $$so{\left(\frac{z-i}{z+i}\right)}^n=cos\theta +isin\theta =e^{i\theta }$$ $$\frac{z-i}{z+i}=e^{i\frac{\theta }{n}}\iff z-i-(z+i)e^{i\frac{\theta }{n}}=0\iff z(1-e^{i\frac{\theta }{n}})=i(1+e^{i\frac{\theta }{n}})$$ $$z=i\frac{1+e^{i\frac{\theta }{n}}}{1-e^{i\frac{\theta }{n}}}$$

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