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
$\dfrac{1-\cos4\theta+i\sin4\theta}{\sin2\theta+2i\cos^2\theta}=\dfrac{1-(\cos\theta+i\sin\theta)^4}{2\sin\theta\cos\theta+2i\cos^2\theta}=\dfrac{1-(\cos\theta+i\sin\theta)^4}{2\cos\theta(\sin\theta+i\cos\theta)}$
 A: First I'd use some trig identities to get everything in terms of $\sin \theta$ and $\cos \theta$ (or perhaps $\sin 2\theta$ and $\cos 2\theta$).
Next I'd make the denominator purely real by multiplying top and bottom by the complex conjugate of the denominator.
From there you can split the fraction easily into a purely real and purely imaginary part, and go from there.
There may be easier ways to do it, but this will get you there.  Maybe try using Euler's identity; as a start, the numerator is also $1 - e^{-4i\theta}.$
A: I would use the facts that $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$ and $\mathrm{arg}\left(\frac{z_1}{z_2}\right)=\mathrm{arg}(z_1)-\mathrm{arg}(z_2)$.
A: 1 - cos4θ = sin(2θ)^2+cos(2θ)^2 - cos(2θ)^2 + sin(2θ)^2 = 2sin(2θ)^2
sin4θ = 2sin2θcos2θ
from here follows:1 - cos4θ + isin4θ = 2sin2θ(sin2θ+icos2θ) = 4sinθcosθ(sin2θ+icos2θ)
when you do division by 2cosθ(sinθ+icosθ) you are left with 2sinθ(sin2θ+icos2θ)/(sinθ+icosθ)
that means |z|=2sinθ and arg z=θ since  sin2θ+icos2θ has rho = 1 and arg =2θ a sinθ+icosθ has rho = 1 and arg =θ. 2θ - θ = θ
A: 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 $\arg z = \theta$.
