Let $f(z)$ be an analytic function in an open ball $B(z_0,R)=\{z\in \mathbb{C}:|z-z_0|<R\}$ such that $f'(z_0)\neq 0$.

Question

Let $f(z)$ be an analytic function in an open ball $B(z_0,R)=\{z\in \mathbb{C}:|z-z_0|<R\}$ such that $f'(z_0)\neq 0$. Prove that $\frac{2\pi i}{f'(z_0)}=\lim_{r \to 0}\int_{C_r}\frac{1}{f(z)-f(z_0)}dz$, where $C_r$ is the positively oriented circle $|z-z_0|=r$.

Answer 1

Since $f'(z_0)\neq 0$, $\ f$ is locally injective near $z_0$ and so is injective in the ball $B(z_0,r)$, for sufficiently small $r$.

Define $$ g(z)=\frac{z-z_0}{f(z)-f(z_0)} $$ for $z \in B(z_0,r), z\ne z_0$ and $g(z_0)=\dfrac1{f'(z_0)}$.

Then $g$ is holomorphic and we can use Cauchy's integral formula: $$ \int_{C_r}\frac{1}{f(z)-f(z_0)}dz = \int_{C_r}\frac{z-z_0}{f(z)-f(z_0)}\frac{1}{z-z_0}dz = \int_{C_r}\frac{g(z)}{z-z_0}dz = 2\pi i \ g(z_0) = \frac{2\pi i}{f'(z_0)} $$