What is this equal to? : $|A+B|^2$ where $A = P e^{ia}$ and $B = Q e^{ib}$

Question

$A$ and $B$ are two complex numbers:

$A = P e^{ia}$

$B = Q e^{ib}$

I would like to know what is this equal to? :

$|A+B|^2$

Please also give a small proof if possible.

Answer 1

Algebraic proof: $$ A=P(\cos a+i \sin a) \quad \land \quad B=Q(\cos b+i \sin b) \Rightarrow $$ $$ A+B= P\cos a+Q\sin b +i(P\sin a+Q\sin b) \Rightarrow $$ $$ |A+B|^2=(P\cos a+Q\sin b)^2 +(P\sin a+Q\sin b)^2 = $$ $$ =P^2+Q^2+2PQ \cos(a-b) $$ Geometric proof:

$A+B$ is the diagonal of a parallelogram that has sides lenghts $P$ and $Q$. The angle opposite to this diagonal is $\pi-(a-b)$ so, appling cosine rules you have the same result.