# Reciprocal of Holomorphic Function

## Theorem

Let $f: \C \to \C$ be a complex function.

Let $U \subseteq \C$ be an open set such that $f$ has no zeros in $U$.

Suppose further that $f$ is holomorphic in $U$.

Then the complex function

$\displaystyle \frac 1 {f_{\restriction U}} : U \to \C$

is holomorphic.

## Proof

Let $g : U \to \mathbb C$ be such that $g(x) = 1/f(x)$.

Since $f(x)$ is nonzero for $x \in U$, $g$ is well-defined.

By Quotient Rule for Continuous Functions, $g$ is continuous.

Let $z_0 \in U$.

Since $g$ is continuous,

$\displaystyle\lim_{h \to 0} \frac 1 {f(z_0 + h)} = \frac 1 {f(z_0)}.$

Since $f$ is holomorphic,

$\displaystyle\lim_{h \to 0} \frac {f(z_0 + h) - f(z_0)} {h} = f'(z_0).$
 $\displaystyle \displaystyle \lim_{h \to 0} \frac {g(z_0 + h) - g(z_0)} {h}$ $=$ $\displaystyle \displaystyle \lim_{h \to 0} \frac {\frac 1 {f(z_0 + h)} - \frac 1 {f(z_0)} } {h}$ $\displaystyle$ $=$ $\displaystyle -\frac 1 {f(z_0)} \displaystyle \lim_{h \to 0} \left(\left(\frac 1 {f(z_0 + h)}\right)\left(\frac{f(z_0 + h) - f(z_0)}{h}\right) \right)$ $\displaystyle$ $=$ $\displaystyle -\frac 1 {f(z_0)^2} \cdot f'(z_0).$

It follows that $g$ is holomorphic.

$\blacksquare$