Proposition

Suppose $f$ has an isolated singularity at $z_0$.

1. Show that $f’$ has an isolated singularity at $z_0$.
2. Find $\Res_{z = z_0}(f’)$.

Solution

1

Since $f$ has an isolated singularity at $z_0$, there exists a punctured disk with the radius $r$ such that $f$ is holomorhpic in it. Then for every point $z’$ such that $0 < \abs{z_0 - z’} < r$, $f’$ is differentiable at $z’$. Thus $f’$ has an isolated singularity at $z_0$.

2

Let $\sum_{k \in \mathbb{Z}} c_k(z - z_0)^k$ be the Laurent series of $f$. Then $f’ = \sum_{k \in \mathbb{Z}} kc_k(z - z_0)^{k - 1}$. Thus the coefficient of $1/z$ is $0 \cdot c_0 = 0$. Thus $\Res_{z = z_0}(f’) = 0$.