Cauchy's Integral Theorem & Formula
Cauchy's Integral Theorem & Formula
Cauchy's integral theorem states that contour integrals of analytic functions over closed curves in simply connected domains are zero. Cauchy's integral formula then recovers $f(z_0)$ from an integral around any closed curve encircling $z_0$. These results have no real-variable analogue — they show that analytic functions are completely determined by boundary values and are infinitely differentiable, a dramatic strengthening of real differentiability.
Cauchy's Integral Formula
Let $f$ be analytic in a simply connected domain $D$, and let $\gamma$ be a positively oriented simple closed curve in $D$ enclosing $z_0$. Then $$f(z_0)=\frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{z-z_0}\,dz.$$ More generally, $f^{(n)}(z_0)=\frac{n!}{2\pi i}\oint_\gamma\frac{f(z)}{(z-z_0)^{n+1}}\,dz$. Consequence: analytic functions are infinitely differentiable (every holomorphic function is $C^\infty$ and even $C^\omega$ — real-analytic). A power series $\sum a_n(z-z_0)^n$ converges in a disk of radius $R=1/\limsup|a_n|^{1/n}$.
Cauchy's Integral Theorem
If $f$ is analytic on a simply connected domain $D$, then for any closed rectifiable curve $\gamma$ in $D$: $$\oint_\gamma f(z)\,dz=0.$$ Proof (Green's theorem approach): $\oint_\gamma f\,dz=\oint_\gamma(u+iv)(dx+i\,dy)=\oint(u\,dx-v\,dy)+i\oint(v\,dx+u\,dy)$. By Green's: each integral $=\pm\iint(u_y+v_x)\,dA$ and $\iint(u_x-v_y)\,dA$, both zero by CR equations.
Example 1
Evaluate $\oint_{|z|=2}\frac{e^z}{z-1}\,dz$.
Solution: $f(z)=e^z$ is analytic everywhere; $z_0=1$ is inside $|z|=2$. By Cauchy's formula: $\oint\frac{e^z}{z-1}\,dz=2\pi i\cdot f(1)=2\pi i e$.
Example 2
Compute $\oint_{|z|=1}\frac{\cos z}{z^3}\,dz$.
Solution: By the generalized Cauchy formula with $n=2$: $\oint\frac{f(z)}{z^3}\,dz=\frac{2\pi i}{2!}f''(0)$ where $f(z)=\cos z$. $f''(z)=-\cos z$, so $f''(0)=-1$. Integral $=\frac{2\pi i}{2}(-1)=-\pi i$.
Practice
- Compute $\oint_{|z|=2}\frac{z^2}{z^2-1}\,dz$ using partial fractions and Cauchy's formula.
- Prove the maximum modulus principle: if $f$ is analytic and non-constant on a domain, $|f|$ has no interior maximum.
- Show that Cauchy's integral formula gives the coefficients of the Taylor series of $f$ around $z_0$.
- Evaluate $\oint_{|z|=3}\frac{e^z}{(z-1)^2(z+2)}\,dz$ using partial fractions.
Show Answer Key
1. $\frac{z^2}{z^2-1}=1+\frac{1/2}{z-1}+\frac{1/2}{z+1}$. By Cauchy's integral formula, $\oint\frac{dz}{z-a}=2\pi i$ for $|a|<2$. Both $z=1$ and $z=-1$ are inside $|z|=2$, so the integral $=2\pi i(\frac{1}{2}+\frac{1}{2})=2\pi i$.
2. If $|f(z_0)|=\max_{z\in D}|f(z)|$ for interior $z_0$, then by Cauchy's formula $f(z_0)=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})d\theta$. Taking moduli: $|f(z_0)|\le\frac{1}{2\pi}\int|f|d\theta\le|f(z_0)|$. Equality requires $|f|=|f(z_0)|$ on the circle, and by continuity throughout the domain. So $f$ is constant.
3. By Cauchy's formula: $f(z)=\frac{1}{2\pi i}\oint\frac{f(w)}{w-z}dw$. Expand $\frac{1}{w-z}=\sum_{n=0}^\infty\frac{(z-z_0)^n}{(w-z_0)^{n+1}}$ and identify $a_n=\frac{1}{2\pi i}\oint\frac{f(w)}{(w-z_0)^{n+1}}dw=\frac{f^{(n)}(z_0)}{n!}$.
4. Partial fractions: $\frac{e^z}{(z-1)^2(z+2)}=\frac{A}{z-1}+\frac{B}{(z-1)^2}+\frac{C}{z+2}$. Residue at $z=1$: $\frac{d}{dz}[\frac{e^z}{z+2}]_{z=1}=\frac{2e}{9}$. Residue at $z=-2$: $\frac{e^{-2}}{9}$. Integral $=2\pi i(\frac{2e}{9}+\frac{e^{-2}}{9})$.