I am having trouble with the following exercise:
Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a holomorphic function and suppose that there is a point $z_0 \in \mathbb{C}$ and a real $R > \lvert z_0 \rvert$ such that
- $f(z_0) = 0$
- $f(z) \ne 0$ for all $z \ne z_0$ with $\lvert z \rvert < R$
- $f^\prime(z_0) \ne 0$
Show that there exists a function $H(z)$ that is holomorphic on the open disc of radius $R$ around the origin and satisfies
$$\frac{1}{f(z)} = \frac{1}{f^\prime(z_0)(z-z_0)} + H(z)$$
All I understand from the statement is that $z_0$ is an isolated singularity, but I do not see at all how to derive said identity for $1/f(z)$. Could you please help me?
Thanks to Conrads's Comment here is the solution: Using the Taylor series expansion of $f$ around $z_0$ we have
\begin{equation} H(z) := \frac{1}{f(z)} - \frac{1}{f^\prime(z_0)(z-z_0)} = \frac{1}{f^\prime(z_0)(z-z_0)+(z-z_0)^2G(z)} - \frac{1}{f^\prime(z_0)(z-z_0)} \end{equation}
where $G(z) = \sum_{n \ge 2} f^{(n)}(z) (z-z_0)^{n-2}$ and this is combination of holomorphic functions and thus holomorphic.