Let $f$ be an entire function on $\mathbb{C}$. Let $z_0\in\mathbb{C}$. Let $C=\{z\in\mathbb{C}\mid |z-z_0|=1\}$. Suppose $f(z)\neq f(z_0)$ for any $|z-z_0|\leq 1$, $z\neq z_0$. Given
$f'(z_0)=2$, $f''(z_0)=3$, $\int_C\frac{f'(z)}{f(z)-f(z_0)}dz=2\pi i$, how to evaluate $\int_C\frac{1}{(f(z)-f(z_0))^2}dz$?
Dear Professor, in the following solution, how to expand $ \dfrac{1}{(z-z_0)^2\left(2 + \dfrac{3}{2} (z - z_0) + \ldots\right)^2}$ to be $\dfrac{1}{4 (z - z_0)^2} - \dfrac{3}{8 (z - z_0)} + \ldots$?
The only singularity of $1/(f(z) - f(z_0))^2$ inside or on $C$ is at $z_0$.
Near $z_0$ we have $$f(z) - f(z_0) = (z - z_0) f'(z_0) + \dfrac{(z-z_0)^2}{2} f''(z_0) + \ldots = 2 (z - z_0) + \dfrac{3}{2} (z - z_0)^2 + \ldots$$ so that $$ \dfrac{1}{(f(z) - f(z_0))^2} = \dfrac{1}{(z-z_0)^2\left(2 + \dfrac{3}{2} (z - z_0) + \ldots\right)^2} = \dfrac{1}{4 (z - z_0)^2} - \dfrac{3}{8 (z - z_0)} + \ldots $$ i.e. the residue of $\dfrac{1}{(f(z) - f(z_0))^2}$ at $z=z_0$ is $3/8$. You don't need the information about $\int_C \dfrac{f'(z)}{f(z) - f(z_0)}\; dz$ (but you could compute it...)