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---
title: Cauchy's Integral Formula
parent: One Complex Variable
grand_parent: Complex Analysis
nav_order: 3
# cspell:words
---
# {{ page.title }}
{: .theorem-title }
> {{ page.title }}
>
> Let $f : G \to \CC$ be a function holomorphic in an open subset $G \subset \CC$.
> Let $\gamma$ be a contour in $G$ such that the interior of $\gamma$ is contained in $G$.
> Then for any point $a$ in the interior of $\gamma$,
>
> $$
> f(a) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-a} \, dz.
> $$
> {: .katex-display .mb-0 }
{% proof %}
{% endproof %}
{: .theorem-title }
> {{ page.title }} (Generalization)
> {: #cauchys-integral-formula-generalized }
>
> Let $f : G \to \CC$ be a function holomorphic in an open subset $G \subset \CC$.
> Then the $n$th derivative $f^{(n)}$ exists for every $n \in \NN$.
> If $\gamma$ is a contour in $G$ such that the interior of $\gamma$ is contained in $G$,
> then for any point $a$ in the interior of $\gamma$,
>
> $$
> f^{(n)} (a) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} \, dz.
> $$
> {: .katex-display .mb-0 }
{% proof %}
{% endproof %}
The last formula may be rewritten as
$$
\int_{\gamma} \frac{f(z)}{(z-a)^n} \, dz = \frac{2 \pi i}{(n-1)!} f^{(n-1)}(a)
$$
and is often used to compute the integral.
## Many Consequences
{: .theorem-title }
> Cauchy's Estimate
> {: #cauchys-estimate }
>
> Let $f$ be holomorphic on an open set containing the disc with center $a$ and radius $r>0$.
> Then
>
> $$
> \norm{f^{(n)}(a)} \le \frac{n!}{r^n} \sup_{\abs{z-a} = r} \norm{f(z)} \qquad \forall n \in \NN.
> $$
> {: .katex-display .mb-0 }
{% proof %}
From [{{ page.title }}](#cauchys-integral-formula-generalized)
for the circular contour around $a$ with radius $r$ we obtain
$$
\begin{aligned}
\norm{f^{(n)}(a)} &\le \frac{n!}{2\pi} \sup_{\abs{z-a} = r} \norm{f(z)} \, \int_{\abs{z-a} = r} \frac{dz}{\abs{z-a}^{n+1}}.
\end{aligned}
$$
Note that the supremum is finite (and is attained),
because $f$ is continuous and the circle is compact.
Clearly, the integral evaluates to $2 \pi r / r^{n+1}$
and the right hand side of the inequality reduces to the desired expression.
{% endproof %}
---
Recall that an *entire* function is a holomorphic function that is defined everywhere in the complex plane.
{: .theorem-title }
> Liouville's Theorem
> {: #liouvilles-theorem }
>
> Every bounded entire function is constant.
{% proof %}
Consider an entire function $f$ and assume that $\norm{f(z)} \le M$ for all $z \in \CC$ and some $M > 0$.
Since $f$ is holomorphic on the whole plane, we may make
[Cauchy's Estimate](#cauchys-estimate)
for all disks centered at any point $a \in \CC$ and with any radius $r>0$.
For the first derivative, we have $\norm{f'(a)} \le M/r$, which tends to $0$ for $r \to \infty$.
Hence $f' = 0$ in the whole plane. This
[implies](/pages/complex-analysis/one-complex-variable/basics.html#holomorphic-function-is-constant-if-derivative-vanishes)
that $f$ is constant.
{% endproof %}
---
|
