[Math] argument of $z=(\sin(\theta) + i(1-\cos(\theta))^2$

Question

I am trying to find the argument of $z=(sin(\theta) + i(1-cos(\theta))^2$ for $0 < \theta< \pi/2$ in its simplest form.

I've tried expanding it out:

\begin{align*}
z&=(\sin^2 \theta + 2 i \sin \theta \cos \theta + i^2 (1-\cos \theta)^2)\\
&=-\cos(2 \theta) +2 \cos \theta – 2i \sin \theta – i \sin(2 \theta) – 1 \\
&=-\operatorname{cis}(2 \theta) + 2 \operatorname{cis}(\theta) -1
\end{align*}

so $z$ is a quadratic in a complex number, im not sure if this helps in finding arg(z) though? I'm a bit rusty with my complex numbers

Answer 1

HINT:

$$\sin\theta+i(1-\cos\theta)=2\sin\dfrac\theta2\left(\cos\dfrac\theta2+i\sin\dfrac\theta2\right)$$

Now use de Moivre's formula