Question

$z=cis\theta ,(\theta \ne (2n+1\left)\frac{\pi }{2}\right)$ Given: $z=cis\theta =\cos \theta +\sin \theta$.

Find $\frac{z^{2n}-1}{z^{2n}+1}$.

Answer 1

$\frac{z^{2n}-1}{z^{2n}+1}=\frac{ci{s}^{2n}\theta -1}{ci{s}^{2n}\theta +1}$

$=\frac{cis\left(2n\theta \right)-1}{cis\left(2n\theta \right)+1}$ ($\because ci{s}^a\left(\theta \right)=cis\left(a\theta \right)$)

$=\frac{\cos \left(2n\theta \right)-1+i\sin \left(2n\theta \right)}{\cos \left(2n\theta \right)+1+i\sin \left(2n\theta \right)}\times \frac{\cos \left(2n\theta \right)-1-i\sin \left(2n\theta \right)}{\cos \left(2n\theta \right)-1-i\sin \left(2n\theta \right)}$

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

$=\frac{{\cos }^2\left(2n\theta \right)-2\cos \left(2n\theta \right)+1+{\sin }^2\left(2n\theta \right)}{co{s}^2\left(2n\theta \right)-1+i\sin \left(2n\theta \right)\left(\cos \right(2n\theta )-1-(cos\left(2n\theta \right)+1\left)\right)+{\sin }^2\left(2n\theta \right)}$

$=\frac{2-2\cos \left(2n\theta \right)}{1-1-2i\sin \left(2n\theta \right)}$

$=i\frac{1-\cos \left(2n\theta \right)}{\sin \left(2n\theta \right)}$

$=i\text{cosec}\left(2n\theta \right)-i\cot \left(2n\theta \right)$