Let $\gamma$ be a Jordan Curve and $f$ a holormorphic function in Ext($\gamma$), at $\infty$ and along $\gamma$. Prove that
$\displaystyle\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}dw = f(\infty) - f(z)$ if $ z \in Ext(\gamma)$
and
$\displaystyle\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}dw = f(\infty)$ if $ z \in Int(\gamma)$
i dont know how to proceed, Cauchy Theorem say that
$\displaystyle\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}dw = 0$ if $ z \in Ext(\gamma)$
and
$\displaystyle\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}dw = f(z)$ if $ z \in Int(\gamma)$
if $\gamma$ is any closed curve, so what should i do about that $f(\infty)$?