Q:

Complex Analysis Identity

Prove that for all z,w $\in\mathbb C$
$\frac{Im(z\bar w)}{zw}$ = $\frac{Im(z)}{z}$ - $\frac{Im(w)}{w}$
Any tips to prove this?

A:

Let $z = r_1(\cos \alpha + i \sin \alpha), w = r_2(\cos \beta + i\sin \beta)$. Then we need
$$\frac{\sin(\alpha - \beta)}{e^{i(\alpha+\beta)}} = \frac{\sin\alpha}{e^{i\alpha}} - \frac{\sin \beta}{e^{i\beta}}$$
Thus we need
$$\sin(\alpha - \beta) = (\cos \beta+i\sin \beta)\sin\alpha - (\cos\alpha + i\sin\alpha)\sin \beta$$
Expanding the right side, this follows.

