# Cauchy’s Integral Formula

Cauchy’s Integral Formula

Let $f:G→C$ be a function holomorphic in an open subset $G⊂C$. Let $γ$ be a contour in $G$ such that the interior of $γ$ is contained in $G$. Then for any point $a$ in the interior of $γ$,

$f(a)=2πi1 ∫_{γ}z−af(z) dz.$

Cauchy’s Integral Formula (Generalization)

Let $f:G→C$ be a function holomorphic in an open subset $G⊂C$. Then the $n$th derivative $f_{(n)}$ exists for every $n∈N$. If $γ$ is a contour in $G$ such that the interior of $γ$ is contained in $G$, then for any point $a$ in the interior of $γ$,

$f_{(n)}(a)=2πin! ∫_{γ}(z−a)_{n+1}f(z) dz.$

The last formula may be rewritten as

$∫_{γ}(z−a)_{n}f(z) dz=(n−1)!2πi f_{(n−1)}(a)$and is often used to compute the integral.

## Many Consequences

Cauchy’s Estimate

Let $f$ be holomorphic on an open set containing the disc with center $a$ and radius $r>0$. Then

$∥f_{(n)}(a)∥≤r_{n}n! ∣z−a∣=rsup ∥f(z)∥∀n∈N.$

Proof From Cauchy’s Integral Formula for the circular contour around $a$ with radius $r$ we obtain

$∥f_{(n)}(a)∥ ≤2πn! ∣z−a∣=rsup ∥f(z)∥∫_{∣z−a∣=r}∣z−a∣_{n+1}dz . $Note that the supremum is finite (and is attained), because $f$ is continuous and the circle is compact. Clearly, the integral evaluates to $2πr/r_{n+1}$ and the right-hand side of the inequality reduces to the desired expression. $□$

Recall that an *entire* function is a holomorphic function that is defined everywhere in the complex plane.

Liouville’s Theorem

Every bounded entire function is constant.

Proof Consider an entire function $f$ and assume that $∥f(z)∥≤M$ for all $z∈C$ and some $M>0$. Since $f$ is holomorphic on the whole plane, we may make Cauchy’s Estimate for all disks centered at any point $a∈C$ and with any radius $r>0$. For the first derivative, we have $∥f_{′}(a)∥≤M/r$, which tends to $0$ for $r→∞$. Hence $f_{′}=0$ in the whole plane. This implies that $f$ is constant. $□$