Cauchy's Integral Formula

Cauchy's Integral Formula is a fundamental result in complex analysis. It states that if $U$ is a subset of the complex plane containing a simple counterclockwise loop $C$ and the region bounded by $C$ , and $f$ is a complex-differentiable function on $U$ , then for any $z_0$ in the interior of the region bounded by $C$ , $\frac{1}{2\pi i} \int\limits_C \frac{f(z)}{z- z_0}dz = f(z_0) .$

Proof

Let $D$ denote the interior of the region bounded by $C$ . Let $C_r$ denote a simple counterclockwise circle about $z_0$ of radius $r$ . Since the interior of the region bounded by $C$ is an open set, there is some $R$ such that $C_r \subset D$ for all $r \in (0, R)$ . For such values of $r$ , $\int\limits_C \frac{f(z)}{z-z_0}dz = \int\limits_{C_r} \frac{f(z)}{z-z_0}dz ,$ by application of Cauchy's Integral Theorem.

Since $f$ is differentiable at $z_0$ , for any $\epsilon$ we may pick an arbitarily small $r>0$ such that $\left\lvert \frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) \right\rvert < \epsilon$ whenever $\lvert z - z_0 \rvert \le r$ . Let us parameterize $C_r$ as $h(t) = r e^{it}+ z_0$ , for $t\in [0,2\pi]$ . Since $\int\limits_{C_r} f'(z_0)dz = 0$ (again by Cauchy's Integral Theorem), it follows that $\begin{align*} \biggl\lvert \int\limits_{C_r} \frac{f(z)}{z-z_0}dz - \int\limits_{C_r} \frac{f(z_0)}{z-z_0}dz \biggr\rvert &= \biggl\lvert \int\limits_{C_r} \left[ \frac{f(z) - f(z_0)}{z-z_0} - f'(z_0) \right] dz \biggr\rvert \\ &\le \int\limits_0^{2\pi} \left\lvert \frac{f(h(t)) - f(z_0)}{ h(t) - z_0} - f'(z_0) \right\vert \cdot r dt \\ &< \int\limits_0^{2\pi} \epsilon \cdot r dt = 2\pi \epsilon r . \end{align*}$ Since $\epsilon$ and $r$ can simultaneously become arbitrarily small, it follows that $\begin{align*} \int\limits_C \frac{f(z)}{z-z_0}dz &= \int\limits_{C_r} \frac{f(z_0)} {z- z_0}dz \\ &= f(z_0) \int\limits_{0}^{2\pi} \frac{h'(t)}{h(t) - z_0} dt \\ &= f(z_0) \int\limits_{0}^{2\pi} \frac{ir e^{it}}{re^{it}} dt \\ &= f(z_0) \cdot 2\pi i , \end{align*}$ which is equivalent to the desired theorem. $\blacksquare$

Consequences

By induction, we see that the $n$ th derivative of $f$ at $z_0$ is $f^{(n)}(z_0) = \frac{n!}{2\pi i} \int\limits_C \frac{f(z)}{(z-z_0)^{n+1}}dz,$ for $n>0$ . In particular, the $n$ th derivative exists at $z_0$ , for all $n>0$ . In other words, if a function $f$ is complex-differentiable on some region, then it is infinitely differentiable on the interior of that region.

Since the $(n+1)$ th derivative exists in general, it follows that the $n$ th derivative is continuous. This is not true for functions of real variables! For instance the real function $f(x) = \begin{cases} x \sin(1/x), & x \neq 0 \\ 0, & x=0 \end{cases}$ is everywhere differentiable, but its derivative is mysteriously not continuous at $x=0$ . In complex analysis, the mystery disappears: the function $z\sin(1/z) = z\frac{e^{i/z} - e^{-i/z}}{2i}$ has an essential singularity at $z=0$ , so we can't establish a derivative there in any case.

The theorem is useful for estimating a function (or its $n$ th derivative) at a point based on the behavior of the function around the point. For instance, the theorem yields an easy proof that holomorphic functions are in fact analytic.

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Cauchy's Integral Formula

Proof

Consequences

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