Cauchy's integral formula and its consequences

Cauchy's theorem says that integrals of analytic functions round closed contours vanish. Its companion, the integral formula, says something much more surprising: the values of an analytic function inside a circle are completely determined by its values on the circle. From this one formula flow the deepest properties of analytic functions. They are infinitely differentiable. Their derivatives obey explicit bounds. A bounded entire function is constant, which gives a two-line proof of the fundamental theorem of algebra. And the modulus of a non-constant analytic function has no interior maximum. This lesson proves all of these.

The formula

Cauchy's integral formula. Let \(f\) be analytic on an open set containing a closed disk \(\overline{D}\), and let \(C\) be its boundary circle, positively oriented. Then for every \(z_0\) inside \(C\),

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

Proof. The integrand is analytic except at \(z_0\), so by the deformation principle of lesson 7 we may replace \(C\) by a small circle \(C_\rho\) of radius \(\rho\) about \(z_0\). Write

\[\oint_{C_\rho}\frac{f(z)}{z - z_0}\,dz = f(z_0)\oint_{C_\rho}\frac{dz}{z - z_0} + \oint_{C_\rho}\frac{f(z) - f(z_0)}{z - z_0}\,dz.\]

The first term is \(2\pi i f(z_0)\). In the second, \(|z - z_0| = \rho\), so by the ML inequality its modulus is at most \(\frac{\max_{C_\rho}|f(z) - f(z_0)|}{\rho}\cdot 2\pi\rho = 2\pi\max_{C_\rho}|f(z) - f(z_0)|\), which tends to 0 as \(\rho \to 0\) by continuity of \(f\). But the left side does not depend on \(\rho\), so the second term is exactly 0. \(\square\)

The same proof works for any positively oriented simple closed contour \(C\) with \(f\) analytic on and inside it, and Cauchy stated the formula in 1831, in a memoir written in Turin. The formula is remarkable. It says that an analytic function behaves like a steady-state temperature distribution: fix the values on the boundary, and the values inside are forced. No such statement holds for real differentiable functions, which can be changed inside a disk while left alone on its boundary.

For points outside the contour the integral is 0 instead, by Cauchy's theorem, since then \(\frac{f(z)}{z - z_0}\) is analytic inside. Both cases are covered by the general version: if \(f\) is analytic on a simply connected domain \(D\) and \(\gamma\) is a closed contour in \(D\) not passing through \(z_0\), then \(\frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{z - z_0}\,dz = n(\gamma, z_0)\,f(z_0)\), where \(n(\gamma, z_0)\) is the winding number of lesson 6. A curve that winds twice round \(z_0\) reproduces \(2f(z_0)\), one that winds once clockwise reproduces \(-f(z_0)\), and one that does not wind round it at all gives 0.

Analytic functions are infinitely differentiable

The integral formula expresses \(f(z_0)\) as an integral in which \(z_0\) appears only in the simple factor \(\frac{1}{z - z_0}\). Differentiating under the integral sign suggests

\[f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z - z_0)^{n+1}}\,dz, \qquad n = 0, 1, 2, \dots\]

Theorem. Under the hypotheses of the integral formula, \(f\) has derivatives of all orders inside \(C\), given by this formula.

Proof for \(n = 1\). Let \(d\) be the distance from \(z_0\) to \(C\) and take \(|h| < \frac{d}{2}\). By the integral formula at \(z_0 + h\) and \(z_0\),

\[\frac{f(z_0 + h) - f(z_0)}{h} = \frac{1}{2\pi i}\oint_C\frac{f(z)}{(z - z_0 - h)(z - z_0)}\,dz.\]

Subtracting the claimed value, the difference is \(\frac{h}{2\pi i}\oint_C\frac{f(z)}{(z - z_0 - h)(z - z_0)^2}\,dz\). On \(C\), \(|z - z_0| \ge d\) and \(|z - z_0 - h| \ge \frac{d}{2}\), so if \(|f| \le M\) on \(C\) and \(C\) has length \(L\), the ML inequality bounds this by \(\frac{|h|}{2\pi}\cdot\frac{2M}{d^3}L \to 0\). The higher derivatives follow by induction with the same estimate. \(\square\)

Since \(f'\) is again analytic, so is \(f''\), and so on: an analytic function is infinitely differentiable. In particular the partial derivatives of \(u\) and \(v\) of every order are continuous, which justifies the claim in lesson 4 that the real and imaginary parts are harmonic. Contrast real analysis, where \(x|x|\) has one derivative but not two.

Cauchy's estimates and Liouville's theorem

Apply the derivative formula on the circle \(|z - z_0| = R\), where \(|f| \le M\). The circle has length \(2\pi R\) and \(|z - z_0|^{n+1} = R^{n+1}\), so

\[|f^{(n)}(z_0)| \le \frac{n!\,M}{R^n}.\]

These are Cauchy's estimates. They say that a bound on \(f\) on a circle controls all its derivatives at the center.

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

Proof. Suppose \(|f| \le M\) on the whole plane. For any \(z_0\) and any \(R > 0\), Cauchy's estimate with \(n = 1\) gives \(|f'(z_0)| \le \frac{M}{R}\). Let \(R \to \infty\): \(f'(z_0) = 0\). So \(f' = 0\) everywhere and \(f\) is constant. \(\square\)

The theorem is named after Joseph Liouville, who lectured on it in 1847, though Cauchy had published it in 1844. It explains why \(\sin z\) and \(\cos z\) had to be unbounded in lesson 5: they are entire and not constant.

The fundamental theorem of algebra

Theorem. Every non-constant polynomial \(p\) with complex coefficients has a root in \(\mathbb{C}\).

Proof. Suppose \(p(z) \ne 0\) for all \(z\). Then \(g = \frac{1}{p}\) is entire. Write \(p(z) = a_nz^n + \dots + a_0\) with \(n \ge 1\) and \(a_n \ne 0\). For large \(|z|\), \(|p(z)| \ge |a_n||z|^n - (|a_{n-1}||z|^{n-1} + \dots + |a_0|) \ge \frac{|a_n|}{2}|z|^n\), so \(|g(z)| \to 0\) as \(|z| \to \infty\). In particular \(|g| \le 1\) outside some disk, and \(g\) is bounded on that closed disk because it is continuous there. So \(g\) is a bounded entire function, hence constant by Liouville, and then \(p\) is constant: a contradiction. \(\square\)

Dividing out a root at a time, a polynomial of degree \(n\) factors as \(a_n(z - z_1)\cdots(z - z_n)\), as stated in lesson 1. Gauss gave several proofs from 1799 onward; this one, from analysis rather than algebra, is perhaps the shortest.

Morera's theorem

Cauchy's theorem has a converse. Morera's theorem (Giacinto Morera, 1886): if \(f\) is continuous on a domain \(D\) and \(\oint_{\partial T} f\,dz = 0\) for every triangle \(T\) in \(D\) with its inside, then \(f\) is analytic on \(D\). Proof: on any disk in \(D\), the construction of lesson 7 gives an antiderivative \(F\) with \(F' = f\). Then \(F\) is analytic, so by the theorem above its derivative \(f\) is analytic too. \(\square\) Morera's theorem is the standard tool for showing that limits, sums and integrals of analytic functions are analytic, since vanishing integrals pass to limits easily.

Uniform limits of analytic functions

Morera's theorem has a powerful consequence, due to Karl Weierstrass. Suppose analytic functions \(f_n\) on a domain \(D\) converge to \(f\) uniformly on every closed disk in \(D\). Then \(f\) is analytic on \(D\), and \(f_n' \to f'\) uniformly on every closed disk as well.

Proof. On a closed disk, \(f\) is a uniform limit of continuous functions, hence continuous. For a triangle \(T\) in a disk in \(D\), \(\oint_{\partial T} f_n\,dz = 0\) by Cauchy–Goursat, and the ML inequality gives \(|\oint_{\partial T}(f - f_n)\,dz| \le \max_{\partial T}|f - f_n|\cdot\operatorname{perimeter}(T) \to 0\). So \(\oint_{\partial T} f\,dz = 0\), and Morera's theorem shows \(f\) is analytic. For the derivatives, apply the derivative formula to \(f - f_n\) on a slightly larger circle: by Cauchy's estimate, \(|f'(z) - f_n'(z)| \le \frac{1}{R}\max_{|w - z| = R}|f(w) - f_n(w)| \to 0\). \(\square\)

Nothing like this is true on the real line. The functions \(\frac{\sin(n^2x)}{n}\) converge uniformly to 0, but their derivatives \(n\cos(n^2x)\) do not converge at all; and by the Weierstrass approximation theorem, every continuous function on an interval, differentiable or not, is a uniform limit of polynomials. In the complex plane, uniform convergence preserves analyticity. Lesson 9 applies the theorem to power series.

Mean values and the maximum modulus principle

Take \(C\) to be the circle \(z = z_0 + re^{it}\) in the integral formula. Then \(dz = ire^{it}\,dt\) and \(z - z_0 = re^{it}\), so

\[f(z_0) = \frac{1}{2\pi}\int_0^{2\pi}f(z_0 + re^{it})\,dt.\]

The value at the center is the average of the values on the circle: the mean value property. Taking real parts, harmonic functions have the same property.

Maximum modulus principle. If \(f\) is analytic on a domain \(D\) and \(|f|\) has a local maximum at a point of \(D\), then \(f\) is constant on \(D\).

Proof sketch. Suppose \(|f(z_0)| \ge |f(z)|\) for \(z\) in a disk about \(z_0\). For each small \(r\), \(|f(z_0)| \le \frac{1}{2\pi}\int_0^{2\pi}|f(z_0 + re^{it})|\,dt \le |f(z_0)|\), so equality holds, which forces \(|f| = |f(z_0)|\) on every small circle, hence on a disk. By lesson 4 an analytic function of constant modulus on a disk is constant there, and the identity theorem of lesson 9 extends this to all of \(D\). \(\square\)

Consequently, on a closed bounded region, \(|f|\) attains its maximum on the boundary. For example, the maximum of \(|z^2 + 1|\) on the closed unit disk is attained on the circle, where at \(z = \pm 1\) it equals 2.

import cmath, math

def circle_integral(g, z0=0, r=1.0, n=2048):
    """Trapezoid rule on a circle: spectrally accurate for smooth periodic integrands."""
    s = 0
    for k in range(n):
        t = 2*math.pi*k/n
        z = z0 + r*cmath.exp(1j*t)
        s += g(z) * 1j*r*cmath.exp(1j*t)
    return s * 2*math.pi/n

f = lambda z: cmath.exp(z) * cmath.cos(z)
for z0 in [0.3 + 0.2j, -0.5j, 0.9]:
    rec = circle_integral(lambda z: f(z)/(z - z0), r=2) / (2j*math.pi)
    print(f"z0={z0}: Cauchy formula {rec:.10f}  direct {f(z0):.10f}")

# Derivatives at 0 from the circle |z| = 1: f^(n)(0) = n!/(2 pi i) * integral f/z^(n+1).
for n in range(5):
    d = math.factorial(n) * circle_integral(lambda z: cmath.sin(z)/z**(n+1)) / (2j*math.pi)
    print(f"sin^({n})(0) approx {d.real:+.10f}")

print("integral of sin z / z^4 on |z|=1:", circle_integral(lambda z: cmath.sin(z)/z**4), " -pi*i/3 =", -math.pi*1j/3)

# Mean value property and the maximum modulus on the unit disk for z^2 + 1.
g = lambda z: z*z + 1
mean = sum(g(0.6*cmath.exp(2j*math.pi*k/1000)) for k in range(1000)) / 1000
print("mean of g on |z|=0.6:", mean, " g(0) =", g(0))
best = max((abs(g(r*cmath.exp(1j*t))), r, t) for r in [0.25*i for i in range(5)] for t in [2*math.pi*k/360 for k in range(360)])
print(f"max |z^2+1| on the grid: {best[0]:.4f} at r = {best[1]}")

z0=(0.3+0.2j): Cauchy formula 1.3051816287+0.1826248425j  direct 1.3051816287+0.1826248425j
z0=(-0-0.5j): Cauchy formula 0.9895848834-0.5406126857j  direct 0.9895848834-0.5406126857j
z0=0.9: Cauchy formula 1.5289138119-0.0000000000j  direct 1.5289138119+0.0000000000j
sin^(0)(0) approx -0.0000000000
sin^(1)(0) approx +1.0000000000
sin^(2)(0) approx +0.0000000000
sin^(3)(0) approx -1.0000000000
sin^(4)(0) approx -0.0000000000
integral of sin z / z^4 on |z|=1: (5.562941629376987e-16-1.047197551196594j)  -pi*i/3 = (-0-1.0471975511965976j)
mean of g on |z|=0.6: (0.9999999999999999-1.6691509285848838e-17j)  g(0) = 1
max |z^2+1| on the grid: 2.0000 at r = 1.0

Common mistakes