Cauchy's Integral Formula

In this lesson you’ll meet the formula that reconstructs an analytic function from its boundary values, use it to evaluate contour integrals immediately, and see the four major theorems that fall out of it.

Cauchy’s integral formula. If $f$ is analytic on and inside a positively oriented simple closed contour $C$, and $z_0$ is inside $C$, then

$$f(z_0) = \frac{1}{2\pi i}\oint_C\frac{f(z)}{z-z_0}\,dz$$

Read that again in words: the values of $f$ on the contour determine its value at every interior point. Nothing in real analysis resembles this. A real function on $[a,b]$ can be anything at all in the middle regardless of its endpoint values.

Taking $C$ to be a circle of radius $r$ centred at $z_0$ and parametrising, the $\frac{1}{z-z_0}$ and the $dz$ combine to leave

$$f(z_0) = \frac{1}{2\pi}\int_0^{2\pi}f\big(z_0+re^{it}\big)\,dt$$

which is the mean value property: $f(z_0)$ is the plain average of $f$ over any circle around it. The figure checks this numerically for $f(z)=z^3+2z+4$ at $z_0=1+i$, where $f(z_0)=4+4i$, and all three radii reproduce it to numerical precision.

Deform $C$ to a small circle $C_\rho$ of radius $\rho$ about $z_0$; the deformation is legitimate because the integrand is analytic between the two contours. Then split:

$$\oint_{C_\rho}\frac{f(z)}{z-z_0}dz = f(z_0)\oint_{C_\rho}\frac{dz}{z-z_0} + \oint_{C_\rho}\frac{f(z)-f(z_0)}{z-z_0}dz$$

The first integral is $2\pi i$, the computation from lesson 8. For the second, continuity of $f$ makes $|f(z)-f(z_0)|$ small on a small circle, and the ML inequality bounds the whole integral by roughly $2\pi\rho\cdot\frac{\varepsilon}{\rho} = 2\pi\varepsilon$, which tends to 0. So the second term vanishes and the first gives the formula. ∎

The $2\pi i$ came from $\oint\frac{dz}{z}$, as promised.

Differentiating with respect to $z_0$ under the integral sign gives an infinite family of formulas:

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

This is where the section’s most surprising fact comes from. The right-hand side makes sense for every $n$ as soon as $f$ is analytic, so:

If $f$ is analytic on a domain, then $f$ has derivatives of all orders there, and each is analytic.

One derivative gives you all of them. Compare the real case, where $x^{4/3}$ has one derivative and not two. This single consequence is why complex analysis is so much stronger than real analysis, and it all comes from being able to write derivatives as integrals.

Liouville’s theorem. A bounded entire function is constant.

Proof. Apply the $n=1$ formula on a circle of radius $R$ about any $z_0$. If $|f|\le M$ everywhere, the ML inequality gives

$$|f'(z_0)| \le \frac{1}{2\pi}\cdot\frac{M}{R^2}\cdot 2\pi R = \frac{M}{R}$$

Let $R\to\infty$: $f'(z_0) = 0$ for every $z_0$, so $f$ is constant. ∎

Fundamental theorem of algebra. Every non-constant polynomial has a complex root.

Proof. If $p$ had no root, $1/p$ would be entire. Since $|p(z)|\to\infty$ as $|z|\to\infty$, the function $1/p$ is bounded, hence constant by Liouville, hence $p$ is constant. Contradiction. ∎

Three lines, for a theorem that resisted algebraic proof for centuries. Every proof of it uses some analysis or topology; this is the shortest.

Maximum modulus principle. If $f$ is analytic and non-constant on a domain, $|f|$ has no interior maximum. It follows from the mean value property: a value that was strictly largest could not equal the average of the surrounding values.

Morera’s theorem. The converse of Cauchy’s theorem. If $f$ is continuous on a domain and $\oint_C f\,dz=0$ for every closed contour, then $f$ is analytic. So vanishing integrals are not just a consequence of analyticity, they characterize it.

In practice the formula is most often read from right to left, as a way to evaluate integrals:

$$\oint_C\frac{f(z)}{z-z_0}\,dz = 2\pi i\,f(z_0), \qquad \oint_C\frac{f(z)}{(z-z_0)^{n+1}}\,dz = \frac{2\pi i}{n!}f^{(n)}(z_0)$$

Given an integral, factor the denominator, check which zeros are inside $C$, and identify what plays the role of $f$. That recipe handles most contour integrals you will meet before residues arrive.

Example 1: Evaluate $\displaystyle\oint_{|z|=2}\frac{e^z}{z}\,dz$.

Solution. Write the integrand as $\frac{f(z)}{z-0}$ with $f(z)=e^z$, which is entire, and $z_0=0$ is inside the contour.

$$\oint_{|z|=2}\frac{e^z}{z}\,dz = 2\pi i\,f(0) = 2\pi i\,e^0 = 2\pi i$$

∎

Example 2: Evaluate $\displaystyle\oint_{|z|=3}\frac{z^2}{z-1}\,dz$.

Solution. Here $f(z)=z^2$ and $z_0=1$, which is inside $|z|=3$.

$$2\pi i\,f(1) = 2\pi i\cdot 1 = 2\pi i$$

∎

Around $|z|=\frac12$ instead, the point 1 would be outside and the answer would be 0 by Cauchy’s theorem. Always check whether the singularity is enclosed.

Example 3: Evaluate $\displaystyle\oint_{|z|=1}\frac{\cos z}{z^3}\,dz$.

Solution. Match $\frac{f(z)}{(z-z_0)^{n+1}}$ with $n+1=3$, so $n=2$, $z_0=0$, $f(z)=\cos z$.

$$\oint\frac{\cos z}{z^3}dz = \frac{2\pi i}{2!}f''(0)$$

Since $f''(z) = -\cos z$, we get $f''(0)=-1$, so the integral is

$$\frac{2\pi i}{2}(-1) = -\pi i$$

∎

Example 4: Evaluate $\displaystyle\oint_{|z|=3}\frac{dz}{(z-1)(z-2)}$.

Solution. Both poles are inside. Split by partial fractions:

$$\frac{1}{(z-1)(z-2)} = \frac{-1}{z-1}+\frac{1}{z-2}$$

Each piece is handled by the formula with $f\equiv1$:

$$-2\pi i + 2\pi i = 0$$

∎

Example 5: Evaluate $\displaystyle\oint_{|z-1|=\frac12}\frac{dz}{(z-1)(z-2)}$.

Solution. This contour is a small circle about 1, so only the pole at 1 is enclosed. Write the integrand as $\frac{f(z)}{z-1}$ with

$$f(z) = \frac{1}{z-2}$$

which is analytic on and inside this contour, since 2 is outside. Then

$$2\pi i\,f(1) = 2\pi i\cdot\frac{1}{1-2} = -2\pi i$$

∎

Absorbing the other factor into $f$ is the standard trick, and it works exactly when the other singularity is outside the contour.

Example 6: Apply the mean value property.

$f$ is entire and $f(z) = 3+4i$ at every point of the circle $|z-2|=1$. What is $f(2)$?

Solution. By the mean value property, $f(2)$ is the average of $f$ over that circle, and the average of a constant is that constant:

$$f(2) = 3+4i$$

∎

In fact the identity theorem then forces $f \equiv 3+4i$ everywhere. Constant on a circle means constant, period.

Example 7: Use Liouville’s theorem.

Show that if $f$ is entire and $|f(z)|\le 5|z|$ for all $z$, then $f(z)=cz$ for some constant $c$ with $|c| \le 5$.

Solution. The bound gives $f(0)=0$, so $g(z) = f(z)/z$ has a removable singularity at 0 and extends to an entire function. On $\mathbb{C}$,

$$|g(z)| = \frac{|f(z)|}{|z|}\le 5$$

so $g$ is bounded and entire, hence constant by Liouville. Writing $g\equiv c$ gives $f(z)=cz$, and $|c|\le5$. ∎

Growth bounds pin down entire functions completely, and this style of argument generalizes: $|f(z)| \le M|z|^n$ forces $f$ to be a polynomial of degree at most $n$.

Numerical differentiation. Computing $f^{(n)}$ by the integral formula, with the circle discretized by the trapezoid rule, is far more accurate than finite differences and does not suffer catastrophic cancellation. It is the standard method for high-order derivatives of analytic functions.

Matrix functions. Defining $e^A$, $\log A$, or $\sqrt{A}$ for a matrix $A$ is done by the Cauchy integral formula with $A$ in place of $z_0$, integrating around a contour enclosing the spectrum. This is how modern libraries compute matrix exponentials for stiff differential equations.

Boundary element methods. Solving a PDE from boundary data alone is possible because the solution inside is determined by the boundary, which for Laplace’s equation is exactly this formula. It reduces a 2D problem to a 1D one.

Eigenvalue counting. Contour-integral eigensolvers count and locate eigenvalues in a region by integrating a resolvent around its boundary. Large-scale electronic structure calculations use this.

Filter reconstruction. Recovering a system’s response inside a region of the complex plane from measurements on a contour is the same reconstruction problem, used in system identification.

Tomography, by analogy. The idea that interior values are determined by boundary measurements is the organizing principle of imaging, and complex analysis provides the cleanest exact instance of it.