Question

If $\pi <\theta <2\pi$ and $z=1+\cos \theta +i\sin \theta$, then write the value of $\mid z\mid$.

Answer 1

$\because z=1+\cos \theta +i\sin \theta$

$\Rightarrow \mid z\mid =\sqrt{(1+\cos \theta {)}^2+{\sin }^2\theta }$

$\Rightarrow \mid z\mid =\sqrt{1+2\cos \theta +{\cos }^2\theta +{\sin }^2\theta }$

$\Rightarrow \mid z\mid =\sqrt{1+2\cos \theta +1}$

$\Rightarrow \mid z\mid =\sqrt{2+2\cos \theta }$

$\Rightarrow \mid z\mid =\sqrt{2}\sqrt{1+\cos \theta }$

$\Rightarrow \mid z\mid =\sqrt{2}\sqrt{2{\cos }^2\frac{\theta }{2}}$

$\Rightarrow \mid z\mid =\sqrt{2}\times \sqrt{2}\left|\cos \frac{\theta }{2}\right|$

$\Rightarrow \mid z\mid =2\left|\cos \frac{\theta }{2}\right|$

$\Rightarrow \mid z\mid =-2\cos \frac{\theta }{2}$

$[\because \pi <\theta <2\pi \Rightarrow \cos \frac{\theta }{2}<0]$

$\therefore \mid z\mid =-2\cos \frac{\theta }{2}$