Question

If the argument of complex number $\sin \theta +i(1-\cos \theta ),0<\theta <\pi$ is $\frac{\theta }{k},k\in R$, then value of $k$ is

Answer 1

Let, $z=\sin \theta +i(1-\cos \theta ),0<\theta <\pi$
Then
$\tan \alpha =\left|\frac{1-\cos \theta }{\sin \theta }\right|\Rightarrow \tan \alpha =\left|\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right|\Rightarrow \tan \alpha =\left|\tan \frac{\theta }{2}\right|=\tan \frac{\theta }{2}(\because \frac{\theta }{2}\in (0,\frac{\pi }{2}))$

For $0<\theta <\pi ,$
$(1-\cos \theta )>0$ and $\sin \theta >0.$
So, $z$ lies in first quadrant.
$\therefore \arg \left(z\right)=\alpha =\frac{\theta }{2}$
So,
$\frac{\theta }{2}=\frac{\theta }{k}\Rightarrow k=2$