Let $f(z)$ be a holomorphic function in $D(0, \rho)$ and let $\gamma(t) = re^it$ with $0 \leq t \leq 2\pi$ and $0 < r < \rho$
I saw the following result: $\int_{\gamma} \dfrac{f(z)}{(z-a)^2}dz = 2 \pi i f'(a)$
I'd like to know where that result comes from.
If you know that holomorphic functions have Taylor series, then this is easy.
Expand $f$ in Taylor series around $z=a$.
Then the Taylor series for $\dfrac{f(z)}{(z-a)^2}$ has the term $\dfrac{f'(a)}{(z-a)}$ and all the other terms have primitives and so integrate to zero.
Therefore, we're left with a single term (hence the name residue): $$ \int_{\gamma} \dfrac{f(z)}{(z-a)^2} dz = \int_{\gamma} \dfrac{f'(a)}{(z-a)} dz = 2 \pi i f'(a) $$
The only technical point is that last integral. You can prove that it is the same as the integral around a small circle centered at $z=a$, which is easy to compute.
The same argument proves that $$ \int_{\gamma} \dfrac{f(z)}{(z-a)^{k+1}} dz = 2 \pi i f^{(k)}(a) $$
