# If the real part of the complex number

Question:

If the real part of the complex number

$(1-\cos \theta+2 \text { isin } \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$, then the value of the integral $\int_{0}^{\theta} \sin x d x$ is equal to :

1. 1

2. 2

3. -1

4. 0

Correct Option: 1

Solution:

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

$=\frac{2 \sin ^{2} \frac{\theta}{2}-2 \mathrm{i} \sin \theta}{(1-\cos \theta)^{2}+4 \sin ^{2} \theta}$

$=\frac{\sin \frac{\theta}{2}-2 \mathrm{i} \cos \frac{\theta}{2}}{4 \sin \frac{\theta}{2}\left(\sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2}\right)}$

$\operatorname{Re}(z)=\frac{1}{2\left(\sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2}\right)}=\frac{1}{5}$

$\sin \frac{2 \theta}{2}+4 \cos ^{2} \frac{\theta}{2}=\frac{5}{2}$

$1-\cos ^{2} \frac{\theta}{2}+4 \cos \frac{\theta}{2}=\frac{5}{2}$

$3 \cos ^{2} \frac{\theta}{2}=\frac{3}{2}$

$\cos ^{2} \frac{\theta}{2}=\frac{1}{2}$

$\frac{\theta}{2}=\mathrm{n} \pi \pm \frac{\pi}{4}$

$\theta=2 \mathrm{n} \pi \pm \frac{\pi}{2}$

$\theta=2 \mathrm{n} \pi \pm \frac{\pi}{2}$

$\theta \in(0, \pi)$

$\theta=\frac{\pi}{2}$

$\int_{0}^{\frac{\pi}{2}} \sin \theta d \theta-[-\cos \theta]_{0}^{\frac{\pi}{2}}$

$=-(0-1)$

$=1$