53  Complex integration

Cauchy’s theorem, the integral formula, and their consequences

53.1 What this chapter helps you do

Symbols to keep handy

These are the bits of notation you'll see a lot. If a line of symbols feels like a fence, read it out loud once, then keep going.

  • _C f(z),dz: contour integral of f along C

Definitions to keep handy

These are the words we keep coming back to. If one feels slippery, come back here and steady it before you push on.

  • contour: A path in the complex plane that you integrate along.

  • contour integral: An integral along a curve; in general the value depends on the path.

  • Cauchy’s theorem: For analytic functions, integrals around closed loops vanish (under the right conditions).

  • Cauchy’s integral formula: A boundary integral that gives the value of an analytic function inside the loop.

  • simply connected: A region with no holes, so loops can be shrunk without crossing a singularity.

Here is the main move this chapter is making, in plain terms. You do not need to be fast. You just need to keep the thread.

  • Coming in: Integration in the real plane is along an interval. Integration in the complex plane is along a curve — a contour. The value generally depends on which path you take. Except when the integrand is analytic: then the path does not matter, and the integral around any closed curve is exactly zero.

  • Leaving with: The integral of an analytic function along any closed curve in a simply connected domain is zero (Cauchy-Goursat theorem). This implies that two paths with the same endpoints give the same integral, and — most powerfully — that knowing an analytic function on a closed contour determines its values everywhere inside (Cauchy’s integral formula). The formula also gives all derivatives: f^(n)(z₀) = (n!/2πi) ∮ f(z)/(z−z₀)^{n+1} dz. This means every analytic function is automatically infinitely differentiable.

53.2 Contours and contour integrals

In real-variable calculus you integrate along a line segment. In complex analysis you integrate along a curve in the plane. The mental model is simple: you walk along a path and keep a running total. What changes is that the path can bend, loop, and circle around features of the function.

A contour (or path) in the complex plane is a piecewise smooth curve C: z = z(t), a \leq t \leq b, where z(t) = x(t) + iy(t) and z'(t) is piecewise continuous. Piecewise smooth means the curve can have finitely many corners, but no cusps or fractal wiggles — at each smooth piece the tangent direction is well-defined.

The contour integral of f along C is:

\int_C f(z)\,dz = \int_a^b f(z(t))\,z'(t)\,dt

How to read the core symbol

  • Symbol: \int_C f(z)\,dz
  • Reads as: “integral of f along the contour C”
  • Means: add up the value of f as you move along the path, accounting for the direction and speed you travel
  • Common misread: this is not “area under a curve”; it is a path-based accumulation

This is a real line integral in disguise. Writing f = u + iv and dz = dx + idy, multiply out the product: (u + iv)(dx + i\,dy) = u\,dx + iu\,dy + iv\,dx + i^2 v\,dy = (u\,dx - v\,dy) + i(v\,dx + u\,dy)

Separating real and imaginary parts:

\int_C f\,dz = \int_C(u\,dx - v\,dy) + i\int_C(v\,dx + u\,dy)

Properties. - Linearity: \int_C (\alpha f + \beta g)\,dz = \alpha\int_C f\,dz + \beta\int_C g\,dz. - Reversal: \int_{-C}f\,dz = -\int_C f\,dz (reversing the direction negates the integral). - Additivity: if C = C_1 + C_2, then \int_C = \int_{C_1} + \int_{C_2}.

53.2.1 The ML inequality

When you need a bound on a contour integral rather than its exact value, the ML inequality is the standard tool:

\left|\int_C f(z)\,dz\right| \leq M \cdot L

where M = \max_{z \in C}|f(z)| and L is the arc length of C. This mirrors the real estimate |\int_a^b f\,dx| \leq \max|f| \cdot (b-a). The application section below shows how to read off M and L in practice.

53.2.2 Integral of 1/z^{n+1} around a circle

\oint_{|z|=r} \frac{dz}{z^{n+1}} = \begin{cases} 2\pi i & n = 0 \\ 0 & n \neq 0 \end{cases}

Proof for n=0: parametrise z = re^{i\theta}, dz = ire^{i\theta}d\theta: \int_0^{2\pi}\frac{ire^{i\theta}}{re^{i\theta}}\,d\theta = \int_0^{2\pi}i\,d\theta = 2\pi i

This single result — that the integral of 1/z around the origin is 2\pi i — is the seed from which the residue theorem grows.

53.3 Cauchy’s theorem

Cauchy’s theorem, in words

If a function is analytic everywhere in a region, then there is no “source” or “vortex” hiding inside. Walking around a closed loop, all the little complex contributions cancel perfectly.

So: no singularities inside ⇒ closed-loop integral is zero.

Cauchy-Goursat theorem. If f is analytic throughout a simply connected domain D (no holes — defined in Chapter 1), then for every closed contour C in D:

\oint_C f(z)\,dz = 0

Intuition. Because the domain has no holes, any closed curve can be continuously shrunk to a point. The integral shrinks with it — to zero.

Consequence: path independence. If f is analytic in a simply connected domain and C_1, C_2 are two paths from z_1 to z_2 in that domain, then \int_{C_1}f\,dz = \int_{C_2}f\,dz. The integral depends only on the endpoints, not the path.

Deformation principle. If f is analytic everywhere between two contours C_1 and C_2 (with the same orientation), then \int_{C_1}f\,dz = \int_{C_2}f\,dz. Contours can be freely deformed through regions where f is analytic, without changing the integral.

53.4 Cauchy’s integral formula

Cauchy’s integral formula, in words

For an analytic function, the boundary values are not just “related” to the interior. They determine the interior.

So: know f on the loop ⇒ you can compute f(z_0) inside by a boundary integral.

Cauchy’s integral formula. Let f be analytic in and on a simple closed contour C (traversed counterclockwise). Then for any z_0 interior to C:

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

Equivalently:

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

Proof sketch. Because f is analytic everywhere inside C except at z_0, the deformation principle applies to the annular region between C and any circle around z_0: the integrand f(z)/(z-z_0) is analytic there. Shrink C to a small circle C_\varepsilon of radius \varepsilon around z_0 without changing the integral. As \varepsilon \to 0, f(z) \to f(z_0) uniformly on C_\varepsilon, and \oint_{C_\varepsilon} dz/(z-z_0) = 2\pi i.

Significance. The value of an analytic function at any interior point is completely determined by its values on the boundary. This has no analogue in real analysis.

53.4.1 Derivatives of all orders

Differentiating the integral formula with respect to z_0 under the integral sign (valid because f is analytic and the denominator never vanishes on C):

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

Rearranged to evaluate a given integral: \displaystyle\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz = \frac{2\pi i}{n!}\,f^{(n)}(z_0).

Consequence. An analytic function has derivatives of all orders, all of which are themselves analytic. In real analysis, once-differentiable functions need not be twice differentiable. In complex analysis, once differentiable means infinitely differentiable — a qualitative difference.

53.5 Major consequences

53.5.1 Liouville’s theorem

A bounded entire function is constant. (An entire function is one that is analytic everywhere in \mathbb{C} — no singularities at all.)

Proof. If |f(z)| \leq M everywhere, the ML inequality applied to the derivative formula gives |f'(z_0)| \leq M/R for any circle of radius R. Letting R \to \infty gives f'(z_0) = 0.

53.5.2 Fundamental theorem of algebra

Every non-constant polynomial p(z) of degree n \geq 1 has exactly n complex roots (counting multiplicity).

Proof sketch. If p had no roots, 1/p(z) would be entire (no zeros means no poles) and bounded (since |p(z)| \to \infty as |z| \to \infty, so |1/p(z)| \to 0), contradicting Liouville’s theorem.

53.5.3 Morera’s theorem

If f is continuous on a domain D and \oint_C f\,dz = 0 for every closed triangle C in D, then f is analytic in D. (The converse of Cauchy’s theorem.)

53.6 Application: real integrals via contour integration

The general strategy — closing a real integral with a contour in the complex plane and applying Cauchy’s theorem — is developed fully in the next chapter. This example shows the technique in its simplest form.

Example. \displaystyle\int_{-\infty}^{\infty} \frac{dx}{1+x^2}.

The integrand f(z) = 1/(1+z^2) has poles at z = \pm i. Close the real-line integral with a large semicircle \Gamma_R of radius R in the upper half-plane, forming a closed contour C = [-R,R] + \Gamma_R.

The semicircle contribution vanishes. On \Gamma_R, |z| = R, so |1+z^2| \geq R^2 - 1 for large R. The ML inequality gives: \left|\int_{\Gamma_R}\frac{dz}{1+z^2}\right| \leq \frac{1}{R^2-1}\cdot \pi R \to 0 \quad\text{as } R\to\infty

Applying Cauchy’s integral formula. Factor the denominator: 1/(1+z^2) = [1/(z+i)]/(z-i). This has the form f(z)/(z-z_0) with f(z) = 1/(z+i) and z_0 = i. Since z_0 = i lies inside the upper half-plane contour and f(z) = 1/(z+i) is analytic there:

\oint_C \frac{dz}{1+z^2} = \oint_C \frac{dz}{(z-i)(z+i)} = 2\pi i \cdot f(i) = 2\pi i \cdot \frac{1}{2i} = \pi

Since the semicircle contributes nothing, the real-line integral equals \pi — confirming the known result \arctan(x)\big|_{-\infty}^{\infty} = \pi.

This calculation uses Cauchy’s integral formula because the integrand has a single simple pole and a purely analytic numerator. In engineering practice, the same closing-contour technique is applied via the residue theorem (Chapter 3), which handles poles of any order and multiple singularities simultaneously. The present example is the simplest special case of that general method.

53.7 Exercises

53.7.1 Exercise 1: Direct computation of a contour integral

Compute \displaystyle\int_C \bar{z}\,dz where C is the straight line from 0 to 1+i.

53.7.2 Exercise 2: Cauchy’s theorem — zero integral

Show that \displaystyle\oint_C \frac{e^z}{z^2+4}\,dz = 0 where C is the circle |z| = 1.

53.7.3 Exercise 3: Cauchy’s integral formula

Evaluate \displaystyle\oint_C \frac{e^z}{z - 1}\,dz where C is the circle |z| = 2 (counterclockwise).

53.7.4 Exercise 4: Derivative formula

Evaluate \displaystyle\oint_C \frac{\cos z}{(z - \pi)^2}\,dz where C: |z| = 4 (counterclockwise). The integrand has a pole at z = \pi inside C, so Cauchy’s theorem alone cannot give zero — you need the derivative formula.

53.7.5 Exercise 5: ML inequality estimate

Without evaluating it exactly, show that \displaystyle\left|\oint_C \frac{dz}{z^3}\right| \leq \frac{\pi}{4}, where C is the upper semicircle |z| = 2 from z = 2 to z = -2.