I have the following exercise that I'm having some trouble solving:
Let $f:U\to\mathbb C$ be an holomorphic function, and let $f'(z_0) \neq 0,z_0 \in U$. Prove that, if $\gamma=z_0+re^{it}$ with $t \in [0,2\pi]$, then, for a sufficiently small amount of $r$, we have that:
$$\frac{2\pi i}{f'(z_0)}=\oint_\gamma \frac{dz}{f(z) – f(z_0)}$$
And then They gave the following tip:
Tip: Notice that the function
$$g(z)=\frac{z – z_0}{f(z) – f(z_0)}$$
has a removable singularity in $z = z_0$
My attempt:
I checked and indeed the function $g(z)$ has a removable singularity in $z = z_0$. This means that: $\text{Res}(g,z_0) = 0$ and because of the residue theorem we have that:
$$\oint_\gamma \frac{z – z_0}{f(z) – f(z_0)} = 2 \pi i \text{Res}(g,z_0)\text{Ind}_\gamma (z_0)$$
Giving us:
$$\oint_\gamma \frac{z – z_0}{f(z) – f(z_0)} = 0$$
This integral is similar to the formula we are trying to prove but I don't know where to go from here. Maybe we need to find some other thing that is equal to $0$ and then equal that some other thing to $\oint_\gamma \frac{z – z_0}{f(z) – f(z_0)}$. How should I continue the proof?
Best Answer
Let $\bar{g}$ be defined by $\bar{g}(z_0)=1/f'(z_0)$ and $\bar{g}(z) = g(z)$ for $z \neq z_0.$ Note that $\bar{g}$ is holomorphic in an neighborhood of $z_0$. Therefore we can use Cauchy's integral formula: $$ \frac{1}{f'(z_0)} = \bar{g}(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{\bar{g}(z)}{z-z_0} dz = \frac{1}{2\pi i} \oint_\gamma \frac{1}{f(z)-f(z_0)} dz $$