Question

If $z=e^{i\theta }$ then find $\frac{z^2-1}{z^2+1}$

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

Answer 1

Answer: (B)

$i\tan \theta$

$\frac{z^2-1}{z^2+1}=\frac{e^{i2\theta }-1}{e^{i2\theta }+1}$
$=\frac{\cos 2\theta +i\sin 2\theta -1}{\cos 2\theta +i\sin 2\theta +1}$
$=\frac{i\sin 2\theta -(1-\cos 2\theta )}{i\sin 2\theta +(1+\cos 2\theta )}$
$=\frac{i\sin 2\theta -2{\sin }^2\theta }{i\sin 2\theta +2{\cos }^2\theta }$
$=\frac{i.2\sin \theta \cos \theta -2{\sin }^2\theta }{i.2\sin \theta \cos \theta +2{\cos }^2\theta }$
$=\frac{2\sin \theta (i\cos \theta -\sin \theta )}{2\cos \theta (i\sin \theta +\cos \theta )}$
$=i\tan \theta \left(\frac{\cos \theta +i\sin \theta }{\cos \theta +i\sin \theta }\right)$
$=i\tan \theta$