Question:
If π < θ < 2π and z = 1 + cos θ + i sin θ, then write the value of
Solution:
$\pi<\theta<2 \pi$
$\frac{\pi}{2}<\frac{\theta}{2}<\pi$ (Dividing by 2 )
$z=1+\cos \theta+$ isin $\theta$
$\Rightarrow|z|=\sqrt{(1+\cos \theta)^{2}+\sin ^{2} \theta}$
$\Rightarrow|z|=\sqrt{1+\cos ^{2} \theta+2 \cos \theta+\sin ^{2} \theta}$
$\Rightarrow|z|=\sqrt{1+1+2 \cos \theta}$
$\Rightarrow|z|=\sqrt{2(1+\cos \theta)}$
$\Rightarrow|z|=\sqrt{2 \times 2 \cos ^{2} \frac{\theta}{2}}$
$\Rightarrow|z|=2 \sqrt{\cos ^{2} \frac{\theta}{2}}$
$\Rightarrow|z|=-2 \cos \frac{\theta}{2} \quad\left[\right.$ Since $\frac{\pi}{2}<\frac{\theta}{2}<\pi, \cos \frac{\theta}{2}$ is negative $]$