calculus - Find modulus and argument of $omega = {frac {sin (P + Q) + i (1 - cos (P + Q))} {(cos P + cos Q) + i (sin P - sin Q) }}$

A past examination paper had the following question that I found somewhat difficult. I tried having a go at it but haven't come around with any possible double angle identities. How would one go about tackling it?

Given:

$$\omega = {\frac {\sin (P + Q) + i (1 - \cos (P + Q))} {(\cos P + \cos Q) + i (\sin P - \sin Q) }}$$

To prove:

$$|\omega| = \tan \frac {P + Q} {2} \qquad\text{and}\qquad \arg(\omega) = Q$$

A guideline on how/ which identity to use would be greatly appreciated.

To give an idea how one would start it is by;

Proof:

$$|\omega| = {\frac {\sqrt{\sin^2 (P + Q) + (1 - \cos (P + Q))^2}} {\sqrt{(\cos P + \cos Q)^2 + (\sin P - \sin Q)^2 }}}$$

I'm still unsure about the above or how the square root come about

Answer

We have
$$\begin{align} N & := \sin^2(P+Q) + (1-\cos(P+Q))^2 = \sin^2(P+Q) + \cos^2(P+Q) + 1 - 2\cos(P+Q) \\ & = 2 (1-\cos(P+Q)) = 2\cdot2\sin^2\frac{P+Q}{2} = 4\sin^2\frac{P+Q}{2} \end{align}$$
and

$$\begin{align} D & = \cos^2P +\cos^2Q + \sin^2P + \sin^2Q + 2(\cos P\cos Q - \sin P \sin Q) \\ &= 2 +2(\cos(P+Q)) = 2(1+\cos(P+Q)) = 4\cos^2\frac{P+Q}{2} \end{align}$$

Now, $$|\omega| = \sqrt{\frac{N}{D}} = \tan\frac{P+Q}{2}$$