Cauchy’s Integral Formula

Cauchy’s Integral Formula

Let $f:G\to \mathbb{C}$ be a function holomorphic in an open subset $G\subset \mathbb{C}$. 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 $\gamma$,

$$f(a)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(z)}{z-a}dz.$$

Cauchy’s Integral Formula (Generalization)

Let $f:G\to \mathbb{C}$ be a function holomorphic in an open subset $G\subset \mathbb{C}$. Then the $n$th derivative $f^{(n)}$ exists for every $n\in \mathbb{N}$. 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 $\gamma$,

$$f^{(n)}(a)=\frac{n!}{2\pi i}{\int }_{\gamma }\frac{f(z)}{(z-a{)}^{n+1}}dz.$$

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

Cauchy’s Estimate

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

$$\Vert f^{(n)}(a)\Vert \le \frac{n!}{r^n}\underset{|z-a|=r}{\sup }\Vert f(z)\Vert \forall n\in \mathbb{N}.$$

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

$$\begin{array}{rl}\Vert f^{(n)}(a)\Vert & \le \frac{n!}{2\pi }\underset{|z-a|=r}{\sup }\Vert f(z)\Vert {\int }_{|z-a|=r}\frac{dz}{|z-a{|}^{n+1}}.\end{array}$$

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. $\square$

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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 $\Vert f(z)\Vert \le M$ for all $z\in \mathbb{C}$ 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 $a\in \mathbb{C}$ and with any radius $r>0$. For the first derivative, we have $\Vert f^{\prime }(a)\Vert \le M/r$, which tends to $0$ for $r\to \infty$. Hence $f^{\prime }=0$ in the whole plane. This implies that $f$ is constant. $\square$

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