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

$\dfrac 1 {f_{\restriction U} } : U \to \C$

is holomorphic.


Proof

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

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

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

Let $z_0 \in U$.

As $g$ is continuous:

$\displaystyle \lim_{h \mathop \to 0} \frac 1 {\map f {z_0 + h} } = \frac 1 {\map f {z_0} }$

As $f$ is holomorphic:

$\displaystyle \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} - \map f {z_0} } h = \map {f'} {z_0}$

By the Combination Theorem for Limits of Functions,

\(\displaystyle \lim_{h \mathop \to 0} \frac {\map g {z_0 + h} - \map g {z_0} } h\) \(=\) \(\displaystyle \lim_{h \mathop \to 0} \frac {\frac 1 {\map f {z_0 + h} } - \frac 1 {\map f {z_0} } } h\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 {\map f {z_0} } \lim_{h \mathop \to 0} \paren {\paren {\frac 1 {\map f {z_0 + h} } } \paren {\frac {\map f {z_0 + h} - \map f {z_0} } h} }\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 {\map f {z_0}^2} \cdot \map {f'} {z_0}\)

It follows that $g$ is holomorphic.

$\blacksquare$