# Complex numbers, de Moivre's theorem

I don't know how I should solve this question:

If $0<\theta<\pi/2$ and $z=(\sin\theta+i(1-\cos\theta))^2$ find in its simplest form $\arg z$

I know that using de Moivre's theorem I should multipy angle by two but there is $i(1-\cosθ)$ and I don't know if I should multipy only $\cos\theta$ which would give $i-i\cos2\theta$ ?

• Just because something is called $\theta$, it does not mean that it is the angle you are looking for. De Moivre's formula says that there is exactly a real number $\alpha\in[0,2\pi)$ such that $\frac{z}{\sqrt{ z\overline z}}=\cos\alpha+i\sin\alpha$. For such real number $\alpha$, it is true that $z^2=(z\overline z)\left(\cos(2\alpha)+i\sin(2\alpha)\right)$. Doing the aforementioned calcs, it might turn out that $\alpha$ has a non-trivial dependence on $\theta$.
– user228113
May 29 '17 at 17:05

$\arg z=\theta$.
• $2\sin\frac{\theta}{2}$ is a common factor. So we have $\displaystyle 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+i\left(2\sin^2\frac{\theta}{2}\right) =2\sin\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)$ May 29 '17 at 17:22