I have to prove that. $$\oint_C \frac{f'(z)}{z-z_0}dz = \oint_C \frac{f(z)}{(z-z_0)^2}dz$$
I found a similar question here: Prove this Cauchy integral $\oint_C \frac{f'(z)}{z-z_0}dz = \oint_C \frac{f(z)}{(z-z_0)^2}dz$
However, I still have some confusions.
If I do $$f'(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z-z_0)^2}dz$$
on the other hand, $$f'(z_0) = \frac{d}{dz_0}f(z_0)$$ I'm not sure how to get $f(z_0)$. I thought $$f(z_0) = \frac{1}{2\pi i}\oint_C\frac{f(z)}{z-z_0}$$ but, it don't seems to work like that.
I'm trying to find $f(z_0)$ using cauchy's formula. I'm not sure if this is the right way to find $f(z_0)$.
$$\oint_C \frac{f(z)}{(z-a)^n} = \frac{2\pi i}{(n-1)!} \frac{d^{n-1}f(z)}{dz^{n-1}}$$
thus, $$f(z_0) = \frac{f'(z)}{1}$$ which is obviously not right.