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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.
resolveinsideC(z−iz+i)n+(z+iz−i)n=2cosθand0<θ<π.ninteger.
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)) ))
(z−iz+i)n+(z−iz+i)−n−2cosθ=0(z−iz+i)2n+1−2(z−iz+i)ncosθ=0((z−iz+i)n)2−2(z−iz+i)ncosθ+(cosθ)2+sin2θ=0((z−iz+i)n−cosθ)2+sin2θ=0(z−iz+i)n−cosθ=(−sin2θ)=+−isinθ=isinθ(because0⩽θ⩽π)so(z−iz+i)n=cosθ+isinθ=eiθz−iz+i=eiθn⇔z−i−(z+i)eiθn=0⇔z(1−eiθn)=i(1+eiθn)z=i1+eiθn1−eiθn

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