Show that if $\displaystyle f$ is analytic within an on a simple closed contour C and $\displaystyle z_{0}$ is not on C, then

$\displaystyle \int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

I've broken this down into two cases: $\displaystyle z_{0}$ is interior to C and $\displaystyle z_{0}$ is exterior to C

Case 1:

Since $\displaystyle z_{0}$ is interior to C we may apply the Cauchy Integral Formula so

$\displaystyle f'(z_{0})*2\pi\\i = \int_{C} \frac{f'(z)}{z-z_{0}} $

and

$\displaystyle f'(z_{0})*2\pi\\i = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

and obviously:

$\displaystyle \int_{C} \frac{f'(z)}{z-z_{0}} = \int_{C} \frac{f(z)}{(z-z_{0})^2}$

However I have no idea what to do when $\displaystyle z_{0}$ is exterior to C. We're only given that f is analytic in the closure of C. I was hoping to use contour deformation or the Cauchy-Goursat Theorem but i don't know how.