Simplification of Complex Number. I would appreciate any hints for the following problem:
Given that
$z=\dfrac{1-\cos4\theta+i\sin4\theta}{\sin2\theta+2i\cos^2\theta}$
show that $\vert z\vert=2\sin\theta$ and arg $z=\theta$
Update:
Using John's suggestion I now have
$\dfrac{1-\cos4\theta+i\sin4\theta}{\sin2\theta+2i\cos^2\theta}=\dfrac{1-(\cos\theta+i\sin\theta)^4}{2\sin\theta\cos\theta+2i\cos^2\theta}=\dfrac{1-(\cos\theta+i\sin\theta)^4}{2\cos\theta(\sin\theta+i\cos\theta)}$

I'm assuming $0 \le \theta < \pi/2$. Since \begin{align}(1 - \cos 4\theta) + i\sin 4\theta &= 2\sin^2 2\theta + i(2\sin 2\theta \cos 2\theta)\\
& = 2\sin 2\theta (\sin 2\theta + i \cos 2\theta)\\
& = 2\sin 2\theta e^{i(2\theta - \pi/2)}
\end{align}
and 
$$\sin 2\theta + 2i\cos^2 \theta = 2\sin \theta \cos \theta + 2i\cos^2 \theta = 2\cos \theta(\sin \theta + i \cos \theta) = 2\cos \theta e^{i(\theta- \pi/2)}$$ 
we have $$z = \frac{2\sin 2\theta}{2\cos \theta}e^{i[(2\theta - \pi/2) - (\theta - \pi/2)]} = (2\sin \theta) e^{i\theta}$$ Therefore $|z| = 2\sin \theta$ and $\arg z = \theta$.
