Cauchy's Integral Formula

Recover a holomorphic function and its derivatives from boundary values.

The idea

Theorem (Cauchy's integral formula).

Let $C$ be a positively oriented simple closed contour, let $f$ be holomorphic on an open set containing $C$ and the region it encloses, and let $a$ be a point inside $C$. Then $f(a) = \frac{1}{2\pi i} \oint_{C} \frac{f(z)}{z - a}\, dz,$ and more generally, for every integer $n \ge 0$, $f^{(n)}(a) = \frac{n!}{2\pi i} \oint_{C} \frac{f(z)}{(z - a)^{n+1}}\, dz.$

The first equation says that the values of $f$ along the boundary determine its value at every point of the interior. Nothing comparable holds for real differentiable functions, whose values at the ends of an interval say almost nothing about the middle.

The formula follows from Cauchy's integral theorem. The integrand $\frac{f(z)}{z - a}$ is holomorphic on the enclosed region except at $a$, so we can deform the contour down to a small circle around $a$ without changing the integral. On a small enough circle $f$ is nearly the constant $f(a)$, which pulls out of the integral and leaves $f(a)\oint \frac{dz}{z - a}$; that last integral, around a circle centered at $a$, equals $2\pi i$.

The second equation is the first differentiated with respect to $a$ under the integral sign, $n$ times. It follows that a function differentiable once in the complex sense has derivatives of every order, since the right-hand side makes sense for every $n$.

Ways to work on it

• Walkthrough. The value formula, when it vanishes, and the derivative version.
• Practice. Read a contour integral off as 2π i f(a).
• Hardest. Evaluate a contour integral with a repeated factor in the denominator.

Not sure where to start? Take the ten-question placement test.