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

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