# Reciprocal of Holomorphic Function

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## 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,

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

By the Combination Theorem for Limits of Functions,

\(\displaystyle \displaystyle \lim_{h \to 0} \frac {g(z_0 + h) - g(z_0)} {h}\) | \(=\) | \(\displaystyle \displaystyle \lim_{h \to 0} \frac {\frac 1 {f(z_0 + h)} - \frac 1 {f(z_0)} } {h}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\frac 1 {f(z_0)} \displaystyle \lim_{h \to 0} \left(\left(\frac 1 {f(z_0 + h)}\right)\left(\frac{f(z_0 + h) - f(z_0)}{h}\right) \right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\frac 1 {f(z_0)^2} \cdot f'(z_0).\) |

It follows that $g$ is holomorphic.

$\blacksquare$