Question

If $z=\cos \theta +i\sin \theta$, then $\frac{z^{2n}-1}{z^{2n}+1}=$

• A. $-\cot n\theta$
• B. $i\tan n\theta$
• C. $\tan n\theta$
• D. $\cot n\theta$

Answer 1

Answer: (B)

$i\tan n\theta$

We have $\frac{z^{2n}-1}{z^{2n}+1}=\frac{{(\cos \theta +i\sin \theta )}^{2n}-1}{{(\cos \theta +i\sin \theta )}^{2n}+1}=\frac{\cos 2n\theta +i\sin 2n\theta -1}{\cos 2n\theta +i\sin 2n\theta +1}$ (Using De Moivre's theorem)

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

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

$=\frac{i\sin n\theta (\cos n\theta +i\sin n\theta )}{\cos n\theta (\cos n\theta +i\sin n\theta )}=i\tan n\theta$