Deriving the Cauchy integral formula from the residue theorem

Question

I'm currently going through complex analysis, and I'm trying to grasp the concept of the whole residue theorem and so on.

I followed the derivation of the residue theorem from the Cauchy integral theorem, and I think I kind of understand what is going on there.

I thought about whether it's possible to derive the Cauchy integral formula from the residue theorem since I read somewhere that the integral formula is just a special case of the residue theorem.

I tried looking it up somewhere but didn't find anything.

Answer 1

Sure it's a special case, since the residue of $g(z):=\dfrac{f(z)}{z-a}$ at $z=a$ is $f(a)$.

This is fairly trivial, but to elaborate a little, recall that $f$ is holomorphic. So we can write $f(z)=f(a)+f'(a)(z-a)+f''(a)/2(z-a)^2+\dots$, and the residue is easy to get.