Cauchy integral formula

If $f$ is analytic on a simply connected domain $\Omega$ and $C$ is a positively oriented simple closed contour in $\Omega$ enclosing a point $z_0$, then

$f(z_0)=\frac{1}{2\pi i}{\oint }_C\frac{f(z)}{z-z_0}dz.$

The value of an analytic function at any interior point of a contour is fixed by its values on the contour. That’s the whole content, and it’s why the CIF carries most of complex analysis.

Why this is striking

In real analysis, boundary values of a function on an interval don’t determine interior values; you can change interior values without affecting boundary values. In complex analysis, analyticity is rigid enough that boundary values do determine interior values, completely.

This rigidity propagates: analytic functions are infinitely differentiable, equal to their Taylor series, bounded entire $\Rightarrow$ constant, and so on. All of those results follow from the CIF.

Sketch of proof

Deform $C$ to a small circle $C_{\epsilon }$ of radius $\epsilon$ around $z_0$ using the Deformation principle (valid since $f(z)/(z-z_0)$ is analytic on the deformation region $\Omega \setminus \{z_0\}$).

On $C_{\epsilon }$, write $f(z)=f(z_0)+(f(z)-f(z_0))$; the second piece is small by continuity of $f$. So

${\oint }_{C_{\epsilon }}\frac{f(z)}{z-z_0}dz=f(z_0){\oint }_{C_{\epsilon }}\frac{dz}{z-z_0}+{\oint }_{C_{\epsilon }}\frac{f(z)-f(z_0)}{z-z_0}dz=2\pi if(z_0)+(\text{error}).$

The error: $|f(z)-f(z_0)|\to 0$ as $\epsilon \to 0$, the contour length is $2\pi \epsilon$, so by ML estimate the error goes to zero. Taking the limit gives the CIF. ∎

CIF as a computational tool

Rewriting the CIF: if $g$ is analytic on $\Omega$ and $C$ encloses $z_0$ (and no other singularities), then

${\oint }_C\frac{g(z)}{z-z_0}dz=2\pi ig(z_0).$

This is a recipe for computing $\oint f$ when $f=g/(z-z_0)$ has a simple pole inside $C$.

Worked example. ${\oint }_{|z|=2}\frac{e^z}{z-1}dz$.

$g(z)=e^z$ is entire. $z_0=1$ is enclosed by $|z|=2$. By CIF: $2\pi i\cdot e^1=2\pi ie$.

Worked example. ${\oint }_{|z|=1}\frac{\cos z}{z}dz$.

$g(z)=\cos z$ entire, $z_0=0$ enclosed. CIF: $2\pi i\cos 0=2\pi i$.

Worked example: multiple poles. ${\oint }_{|z|=3}\frac{1}{(z-1)(z-2)}dz$.

Two poles, $z=1$ and $z=2$, both inside.

Partial fractions: $\frac{1}{(z-1)(z-2)}=\frac{-1}{z-1}+\frac{1}{z-2}$.

Apply CIF to each: $\oint \frac{-1}{z-1}dz=2\pi i(-1)=-2\pi i$; $\oint \frac{1}{z-2}dz=2\pi i$. Sum: $0$.

Alternatively, split the contour by Deformation principle into a small contour around each pole, then apply CIF. Same answer.

Generalized CIF: derivatives

For any nonnegative integer $n$, under the same hypotheses,

$f^{(n)}(z_0)=\frac{n!}{2\pi i}{\oint }_C\frac{f(z)}{(z-z_0{)}^{n+1}}dz.$

So contour integrals over $f(z)/(z-z_0{)}^{n+1}$ give the $n$-th derivative of $f$ at $z_0$, up to constants.

Sketch. Differentiate the CIF with respect to $z_0$, passing the derivative inside the integral. Each differentiation of $1/(z-z_0)$ with respect to $z_0$ gives $1/(z-z_0{)}^2$, with a sign that combines with $n!$ as you iterate, producing $n!/(z-z_0{)}^{n+1}$. That’s where the $n!$ on the right comes from.

Why differentiation under the integral is legal here: the standard sufficient condition is that the integrand and its $z_0$-derivative are both continuous as functions of $(z_0,z)$ on a region of the form $\{z_0\in U\}\times C$ where $U$ is open and $C$ is the contour, and that the integrand-derivative is uniformly bounded on that region. Both hold: pick a small disk $U$ around $z_0$ separated from $C$ by some positive distance $\delta$, so $|z-z_0|\ge \delta$ for all $(z_0,z)\in U\times C$. Then $|1/(z-z_0{)}^{n+1}|\le 1/{\delta }^{n+1}$, uniformly bounded. The Leibniz integral rule (real-variable version, applied componentwise) finishes the argument.

Implication: analytic functions are infinitely differentiable. The right side of the generalized CIF makes sense for any $n$, which means $f^{(n)}(z_0)$ exists for all $n$. Once-differentiable in $\mathbb{C}$ implies infinitely-differentiable in $\mathbb{C}$, a stark contrast to real analysis where you can have $C^1$ but not $C^2$.

Worked example. ${\oint }_{|z|=2}\frac{e^z}{(z-1{)}^3}dz$.

Use the $n=2$ form: $f(z)=e^z$, $z_0=1$, $f^{(2)}(z_0)=e$. So $\oint =\frac{2\pi i}{2!}\cdot e=\pi ie$.

Consequences in 1–3 lines

What the generalized CIF gives you:

• Cauchy’s estimate: $|f^{(n)}(z_0)|\le n!M/r^n$ where $M$ is a bound on $|f|$ on the circle $|z-z_0|=r$. See Cauchy’s estimate.
• Liouville’s theorem: bounded entire $\Rightarrow$ constant. Take $r\to \infty$ in Cauchy’s estimate.
• Fundamental theorem of algebra: every non-constant polynomial has a root. By Liouville.
• Analyticity ⇒ Taylor series convergence: the radius of convergence of the Taylor series equals the distance to the nearest singularity.

In context

The CIF is the main tool for evaluating closed-contour integrals when the integrand has a pole. The Residue theorem extends it to arbitrary isolated singularities, not just simple poles.