If f is analytic on the simple closed contour C, then $\oint_C(f' (z))/((z-z_0 )dz= \oint_C f(z)/(z-z_0 )^2 dz$

Question

If f is analytic within and on the simple closed contour C and z_0 is a point within C,then $$\oint_C\frac{f' (z)}{z-z_0 }dz= \oint_C \frac{f(z)}{(z-z_0 )^2} dz$$

Is this statement true or false? If true why and if false why?

I tried to use Cauchy's integral formula but I couldn't prove true or false.

Answer 1

Note that $f'$ is analytic also. By Cauchy's Integral Formula, both integrals are equal to $2\pi i \,f'(z_0)$.