Showing residue equals $\frac{\pi f(w_0)}{Im(w_0)}$ Suppose $f:A \to \mathbb{C}$ is analytic on an open set $A$ containing the closed half plane $H=\{z \in \mathbb{C}: Im(z) \ge 0\}$ and that there is a finite constant $M$ with $|f(z)| \le M$ for all $z$ in $H$. Show that if $w_0$ is apoint in $\mathbb{C}$ with $Im(w_0)>0$, then $$\int_{- \infty}^{ \infty} \frac{f(x)}{x^2-(2Re(w_0))x+|w_0|^2}=\frac{\pi f(w_0)}{Im(w_0)}$$
My approach: I factored the denominator. I get $(x-w_0)^2$. Calculating the residue of the integrand, I obtain $Res=f'(x)$. Here's where I get confused because I don't obtain the result when I multiply $2 \pi i *f'(x)$. How do I do approach this correctly?

The factorization of the denominator is wrong. It should be: $(x-w_0)(x-\bar w_0)$. Even so, the residue is also wrong. You get
$$
\operatorname{Res}\limits_{x=w_0} \frac{f(x)}{(x-w_0)(x-\bar w_0)} = \lim_{x\to w_0} \frac{f(x)}{x-\bar w_0} = \frac{f(w_0)}{w_0-\bar w_0} = \frac{f(w_0)}{2i\operatorname{Im} w_0}.
$$
