Question:
If $z=\frac{1}{1-\cos \theta-i \sin \theta}$, then $\operatorname{Re}(z)=$
(a) 0
(b) $\frac{1}{2}$
(c) $\cot \frac{\theta}{2}$
(d) $\frac{1}{2} \cot \frac{\theta}{2}$
Solution:
(b) $\frac{1}{2}$
$z=\frac{1}{1-\cos \theta-i \sin \theta}$
$z=\frac{1}{1-\cos \theta-i \sin \theta} \times \frac{1-\cos \theta+i \sin \theta}{1-\cos \theta+i \sin \theta}$
$\Rightarrow z=\frac{1-\cos \theta+i \sin \theta}{(1-\cos \theta)^{2}-(i \sin \theta)^{2}}$
$\Rightarrow z=\frac{1-\cos \theta+i \sin \theta}{1+\cos ^{2} \theta-2 \cos \theta+\sin ^{2} \theta}$
$\Rightarrow z=\frac{1-\cos \theta+i \sin \theta}{1+1-2 \cos \theta}$
$\Rightarrow z=\frac{1-\cos \theta+i \sin \theta}{2(1-\cos \theta)}$
$\Rightarrow \operatorname{Re}(z)=\frac{(1-\cos \theta)}{2(1-\cos \theta)}=\frac{1}{2}$