Question

If $\alpha =\cos \theta +i\sin \theta ,$ then $\frac{1+\alpha }{1-\alpha }$ is equal to

• A. $\cot \frac{\theta }{2}$
• B. $\cot \theta$
• C. $i\cot \frac{\theta }{2}$
• D. $i\tan \frac{\theta }{2}$

Answer 1

The correct option is C $i\cot \frac{\theta }{2}$

$\alpha =\cos \theta +i\sin \theta$

$\therefore \frac{1+a}{1-a}$

$=\frac{1+\cos \theta +i\sin \theta }{1-\cos \theta -i\sin \theta }$

$=\frac{(1+\cos \theta +i\sin \theta )(1-\cos \theta +i\sin \theta )}{(1-\cos \theta -i\sin \theta )(1-\cos \theta +i\sin \theta )}$

$=\frac{1+\cos \theta +i\sin \theta -\cos \theta -{\cos }^2\theta -i\sin \theta \cos \theta +i\sin \theta +i\sin \theta \cos \theta +i2{\sin }^2\theta }{(1-\cos \theta {)}^2-(i\sin \theta {)}^2}$

$=\frac{1+2i\sin \theta -{\cos }^2\theta -{\sin }^2\theta }{1-2\cos \theta +{\cos }^2\theta +{\sin }^2\theta }$

$=\frac{2i\sin \theta }{2-2\cos \theta }$

$=\frac{1\left(2\sin \theta /2\cos \theta /2\right)}{2{\sin }^2\theta /2}$

$=\frac{i\cos \theta /2}{\sin \theta /2}$

$=i\cot \theta /2$