Question

If $a = e^{i \theta}$ , then $\frac{1 + a}{1-a}$ is equal to

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

Answer 1

The Correct Option is C

We have,
$a=e^{i \theta}$
$=\cos \theta+i \,\sin \,\theta$
Now, $\frac{1+a}{1-a}=\frac{1+(\cos\, \theta+i \sin \,\theta)}{1-(\cos \,\theta+i \,\sin\,\theta)}$
$=\frac{(1+\cos\, \theta)+i\, \sin\, \theta}{(1-\cos \,\theta)-i \,\sin\, \theta} $
$=\frac{2 \,\cos ^{2} \frac{\theta}{2}+i \,2 \,\sin \frac{\theta}{2} \,\cos \frac{\theta}{2}}{2 \,\sin ^{2} \frac{\theta}{2}-i\, 2 \,\sin \frac{\theta}{2} \,\cos \frac{\theta}{2}} $
$=\frac{2\, \cos \frac{\theta}{2}\left[\cos \frac{\theta}{2}+i \,\sin \frac{\theta}{2}\right]}{2 \,\sin \frac{\theta}{2}\left[\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}\right]}$
$=\frac{\cot \frac{\theta}{2}\left[\cos \frac{\theta}{2}+i \,\sin \frac{\theta}{2}\right]}{-i\,\left[\cos \frac{\theta}{2}+i\, \sin \frac{\theta}{2}\right]} $
$ \frac{\cot \frac{\theta}{2}}{-i} $
$= i \cot \frac{\theta}{2}$