Cauchy's Integral Formula

Part 9, Chapter 9: Complex Analysis, Holomorphic Functions

Learning objectives

State Cauchy's theorem (closed-contour integral of a holomorphic function is zero)
Use Cauchy's integral formula to recover $f(z_0)$ from boundary data
Compute residues at simple and higher-order poles
Apply the residue theorem to evaluate closed-contour integrals

The miracle of complex analysis is that local differentiability forces global structure. A holomorphic function on a simply connected region is uniquely determined by its values on the boundary, and any integral of it over a closed loop is zero unless the loop encloses a singularity, in which case the integral is exactly $2\pi i$ times the sum of residues inside. This is the residue calculus, and it is the most efficient computational machinery in all of complex analysis. Real definite integrals that take pages to handle by parts collapse to a single residue computation.

Cauchy's theorem

If $f$ is holomorphic on a simply connected open set $U$ and $\gamma$ is any simple closed contour inside $U$ , then

$\oint_\gamma f(z) \, dz = 0.$ gammaf(z),dz=0.

The proof rests on Green's theorem and the Cauchy-Riemann equations: writing $f \, dz = (u + iv)(dx + i dy)$ , the line integral equals $\iint_D (-v_x - u_y) + i(u_x - v_y) \, dA$ D(−vx−uy)+i(ux−vy),dA, both of whose integrands vanish by CR. The hypothesis that $U$ is simply connected is crucial; on multiply connected domains, integrals over closed loops detect the topology.

Cauchy's integral formula

If $f$ is holomorphic inside and on a positively oriented simple closed contour $\gamma$ and $z_0$ lies inside $\gamma$ , then

$f(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - z_0} \, dz.$ gammafracf(z)z−z_0,dz.

The boundary values of $f$ on $\gamma$ completely determine its interior values. Differentiating under the integral sign yields the formula for derivatives:

$f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n+1}} \, dz.$ gammafracf(z)(z−z_0)n+1,dz.

One striking corollary: holomorphic functions are infinitely differentiable. Real differentiability gives no such bonus, $C^1$ does not imply $C^2$ in real analysis, but in $\mathbb{C}$ a single derivative implies all of them.

The widget above lets you visualize contour integrals when the integrand is real-valued on a real interval (useful for understanding the residue trick for real integrals like $\int_{-\infty}^{\infty} \frac{dx}{1 + x^2} = \pi$ −inftyinftyfracdx1+x2=pi, which arises from a semicircular contour and the residue at $z = i$ ).

The residue theorem

When $f$ has isolated singularities $z_1, \ldots, z_k$ inside $\gamma$ but is holomorphic elsewhere inside and on $\gamma$ ,

$\oint_\gamma f(z) \, dz = 2\pi i \sum_{j=1}^k \operatorname{Res}(f, z_j).$ gammaf(z),dz=2piisumj=1koperatornameRes(f,zj).

The residue $\operatorname{Res}(f, z_j)$ j) is the coefficient $a_{-1}$ −1 in the Laurent expansion of $f$ around $z_j$ j. At a simple pole, $\operatorname{Res}(f, z_j) = \lim_{z \to z_j} (z - z_j) f(z)$ zj)f(z). At a pole of order $m$ ,

$\operatorname{Res}(f, z_j) = \frac{1}{(m-1)!} \lim_{z \to z_j} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_j)^m f(z) \right].$

Where this shows up

Quantum field theory: Feynman propagators $\frac{1}{p^2 - m^2 + i\epsilon}$ are evaluated by closing the contour in the lower or upper half-plane and picking up residues at the $\pm m$ poles. This is how QFT computes any amplitude.
Signal processing: The inverse Laplace and inverse Z-transforms are contour integrals computed by residues. Every step response of a linear control system is, at the level of implementation, a residue sum.
Number theory: The explicit formula for $\pi(x)$ (the prime counting function) uses contour integration of $\zeta'(s)/\zeta(s)$ to extract the location of the zeta zeros into the distribution of primes. Even the Prime Number Theorem proof rests on residue calculus.
Statistical mechanics: Partition functions of lattice models near critical points have analytic structure (poles, branch cuts) whose residues encode universal critical exponents. The conformal-bootstrap programme uses these residues to constrain 2D and 3D phase transitions.
Antenna design: Far-field radiation patterns are computed by contour integrals over wavenumber space; closing the contour and using residues at saddle points gives the dominant radiation directions (the stationary-phase method, a residue cousin).

Pause and think: Why does $\oint_{|z|=2} e^z \, dz = 0$ ∣z∣=2ez,dz=0 even though $e^z$ is huge on the circle? $e^z$ is entire, so by Cauchy's theorem the integral over any closed contour vanishes. Now consider $\oint_{|z|=2} \frac{e^z}{z - 1} \, dz$ ∣z∣=2fracezz−1,dz: the integrand has a simple pole at $z = 1$ (inside $|z|=2$ ), residue $e^1 = e$ , so the integral is $2\pi i e$ .

Try it

Evaluate $\oint_{|z|=3} \frac{dz}{z^2 + 1}$ ∣z∣=3fracdzz2+1. (Hint: the integrand has simple poles at $z = \pm i$ , both inside $|z|=3$ . Find the residues and add.)
Predict first: what is $\oint_{|z|=2} \frac{\cos z}{(z - 1)^2} \, dz$ ∣z∣=2fraccosz(z−1)2,dz? Apply the derivative form of Cauchy's formula with $f(z) = \cos z$ , $n = 1$ , $z_0 = 1$ .
Compute the residue of $f(z) = \frac{e^z}{z^3}$ at $z = 0$ . (Hint: expand $e^z$ in Taylor series and find the coefficient of $1/z$ .)
Use the residue theorem to evaluate the real integral $\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1}$ −inftyinftyfracdxx2+1 by closing the contour in the upper half-plane. (The answer is $\pi$ , obtained from the single residue at $z = i$ .)

A trap to watch for

The residue formula $\operatorname{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z)$ z0)f(z) is correct ONLY at a simple pole. At a pole of order $m \geq 2$ , you must differentiate $m - 1$ times and divide by $(m - 1)!$ . Beginners habitually skip this step on double poles like $\frac{\cos z}{(z - 1)^2}$ and get the wrong answer. Sanity check: at a simple pole, $\operatorname{Res} = a_{-1}$ −1 from the Laurent series; at a double pole, $\operatorname{Res}$ is still $a_{-1}$ −1, but extracting it from $\frac{f(z)}{(z - z_0)^2}$ requires one derivative. Always check the order of the pole first; do not autopilot the simple-pole formula.

What you now know

You can recognize when Cauchy's theorem zeros out an integral, apply the integral formula to extract $f(z_0)$ from boundary data, and compute residues at simple and higher-order poles. The next section turns to power series: Taylor expansions inside disks of analyticity, Laurent expansions around isolated singularities, and the classification of singularities via the negative-power tail.

Mark section complete →

References

Garrity, T. (2002). All the Mathematics You Missed: But Need to Know for Graduate School. Cambridge University Press, §9.4-9.5.

Ahlfors, L. V. (1979). Complex Analysis (3rd ed.). McGraw-Hill, ch. 4.

Stein, E. M., Shakarchi, R. (2003). Complex Analysis. Princeton University Press, ch. 2-3.

Conway, J. B. (1978). Functions of One Complex Variable I (2nd ed.). Springer, ch. 4-5.

Brown, J. W., Churchill, R. V. (2014). Complex Variables and Applications (9th ed.). McGraw-Hill, ch. 4-6.

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