# Solve the following

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) $\frac{1}{2}$

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

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

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

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

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

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

$\Rightarrow \operatorname{Re}(z)=\frac{(1-\cos \theta)}{2(1-\cos \theta)}=\frac{1}{2}$