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

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

• Hint: Double angle formulas. – Did Oct 17 '15 at 12:57

$$\sin\theta+i(1-\cos\theta)=2\sin\dfrac\theta2\left(\cos\dfrac\theta2+i\sin\dfrac\theta2\right)$$
Asumming $0<x<\frac{\pi}{2}$:
$$\arg\left((\sin(x) + i(1-\cos(x))^2\right)=$$ $$\arg\left(\sin^2\left(\frac{x}{2}\right)\left(4\cos(x)+4i\sin(x)\right)\right)=$$ $$\arg\left(\left(2\cos(x)-\cos(2x)-1\right)+\left(2\sin(x)-\sin(2x)\right)i\right)=$$ $$\tan^{-1}\left(\frac{2\sin(x)-\sin(2x)}{2\cos(x)-\cos(2x)-1}\right)=$$ $$\tan^{-1}\left(\tan(x)\right)$$