Cauchy's theorem

Lesson 6 showed that a function with an antiderivative integrates to zero round every closed contour, and that \(\frac{1}{z}\) does not, because its antiderivative \(\log z\) cannot be defined all the way round the origin. Cauchy's theorem is the statement that for analytic functions on a region without holes the integral round every closed contour is zero. It is the central theorem of complex analysis. Everything that follows, the integral formula, the infinite differentiability of analytic functions, Taylor and Laurent series and the residue theorem, is deduced from it. This lesson states it, gives two proofs, and develops its most useful consequence, the freedom to deform contours.

A little history

Cauchy presented his memoir on definite integrals taken between imaginary limits in 1825, in which he showed that the integral of a function between two points does not change when the path is varied, provided the function stays finite and continuous between the paths. His argument used the calculus of variations and assumed what we would now call continuity of the derivative. In 1851 Riemann gave the proof via what is now called Green's theorem. In a celebrated paper of 1884, refined in 1900, Édouard Goursat showed that the hypothesis on the derivative is unnecessary: complex differentiability at every point suffices. Goursat's proof, by repeatedly subdividing a triangle, is the one found in most modern texts.

Statement

Cauchy's theorem. Let \(f\) be analytic on a simply connected domain \(D\). Then for every closed contour \(\gamma\) in \(D\),

\[\oint_\gamma f(z)\,dz = 0.\]

Equivalently, integrals of \(f\) in \(D\) depend only on the endpoints, and \(f\) has an antiderivative on \(D\). Recall from lesson 2 that a domain is simply connected if every closed curve in it can be shrunk continuously to a point within it: informally, it has no holes. Disks, half-planes, the plane, and the slit plane are simply connected; an annulus and the punctured plane are not, and \(\oint_{|z| = 1}\frac{dz}{z} = 2\pi i\) shows that the theorem fails on them.

Proof via Green's theorem

Suppose in addition that \(f'\) is continuous, and that \(\gamma\) is a simple closed contour bounding a region \(R\) inside \(D\). Green's theorem from MATH-209 says \(\oint_\gamma (P\,dx + Q\,dy) = \iint_R (Q_x - P_y)\,dA\). From lesson 6,

\[\oint_\gamma f\,dz = \oint_\gamma (u\,dx - v\,dy) + i\oint_\gamma (v\,dx + u\,dy) = \iint_R (-v_x - u_y)\,dA + i\iint_R (u_x - v_y)\,dA.\]

Both integrands vanish by the Cauchy–Riemann equations, so the integral is 0. \(\square\) This proof is short and makes the role of the Cauchy–Riemann equations transparent: they say exactly that the two real line integrals have zero curl. Its weakness is the extra hypothesis of a continuous derivative, which Goursat removed.

Goursat's proof for triangles

Theorem (Cauchy–Goursat). If \(f\) is analytic on an open set containing a closed triangle \(T\) and its inside, then \(\oint_{\partial T} f\,dz = 0\).

Proof. Let \(I = |\oint_{\partial T} f\,dz|\). Join the midpoints of the sides to cut \(T\) into four congruent triangles, each traversed anticlockwise. The integrals over the interior edges cancel in pairs, because each is traversed once in each direction, so the four boundary integrals add up to the integral over \(\partial T\). Hence at least one of them, call its triangle \(T_1\), has \(|\oint_{\partial T_1} f\,dz| \ge \frac{I}{4}\). Repeat: we obtain nested triangles \(T \supset T_1 \supset T_2 \supset \dots\) with

\[\Big|\oint_{\partial T_n} f\,dz\Big| \ge \frac{I}{4^n}, \qquad \operatorname{perimeter}(T_n) = \frac{p}{2^n}, \qquad \operatorname{diameter}(T_n) \le \frac{d}{2^n},\]

where \(p\) and \(d\) are the perimeter and diameter of \(T\). The nested closed triangles shrink to a single point \(z_0\) in \(T\). Since \(f\) is differentiable at \(z_0\), \(f(z) = f(z_0) + f'(z_0)(z - z_0) + \varepsilon(z)(z - z_0)\) with \(\varepsilon(z) \to 0\). The linear part has an antiderivative and integrates to zero round \(\partial T_n\). Given \(\eta > 0\), choose \(n\) large enough that \(|\varepsilon| < \eta\) on \(T_n\); for \(z\) on \(\partial T_n\), \(|z - z_0| \le \frac{d}{2^n}\). The ML inequality gives

\[\frac{I}{4^n} \le \Big|\oint_{\partial T_n}\varepsilon(z)(z - z_0)\,dz\Big| \le \eta\,\frac{d}{2^n}\,\frac{p}{2^n},\]

so \(I \le \eta\,dp\). As \(\eta\) is arbitrary, \(I = 0\). \(\square\)

The proof uses complex differentiability at a single point in each step, and nothing else: no continuity of \(f'\), no Green's theorem.

From triangles to convex and simply connected domains

On a convex domain \(D\) (one containing the segment between any two of its points), fix \(z_1\) in \(D\) and define \(F(z) = \int_{[z_1, z]} f(w)\,dw\) along the segment. For \(z\) and \(z + h\) in \(D\), the triangle with vertices \(z_1, z, z + h\) lies in \(D\), so by Goursat \(F(z + h) - F(z) = \int_{[z, z + h]} f(w)\,dw\). Then

\[\Big|\frac{F(z + h) - F(z)}{h} - f(z)\Big| = \Big|\frac{1}{h}\int_{[z, z + h]}\big(f(w) - f(z)\big)\,dw\Big| \le \max_{w \in [z, z + h]}|f(w) - f(z)| \to 0\]

by the ML inequality and the continuity of \(f\). So \(F' = f\): \(f\) has an antiderivative on \(D\), and by lesson 6 its integral round every closed contour in \(D\) is 0. This proves Cauchy's theorem for convex domains, in particular for disks. The extension to arbitrary simply connected domains uses the same idea along chains of overlapping disks; we will use the theorem in that generality without giving the topological details.

Logarithms on simply connected domains

Cauchy's theorem settles a question left open in lesson 5. Let \(D\) be a simply connected domain not containing 0. The function \(\frac{1}{z}\) is analytic on \(D\), so it has an antiderivative there: fix \(z_1\) in \(D\), choose any value \(c\) of \(\log z_1\), and set

\[L(z) = c + \int_{z_1}^{z}\frac{dw}{w},\]

integrating along any contour in \(D\). Then \(L' = \frac{1}{z}\). To check that \(L\) is a logarithm, differentiate \(ze^{-L(z)}\): the derivative is \(e^{-L} - zL'e^{-L} = 0\), so \(ze^{-L(z)}\) is constant on the domain \(D\), and at \(z_1\) it equals \(z_1e^{-c} = 1\). So \(e^{L(z)} = z\). A branch of the logarithm exists on every simply connected domain avoiding 0, whatever its shape; a spiral-shaped slit, for instance, works as well as a straight one. The same argument shows that a non-vanishing analytic function \(g\) on a simply connected domain has an analytic logarithm, \(\log g = c + \int\frac{g'}{g}\), and hence analytic square roots \(e^{\frac{1}{2}\log g}\). These facts are used repeatedly in the theory of conformal mapping.

The homotopy form of the theorem

Behind both Cauchy's theorem and the deformation principle is a single statement. Two closed contours \(\gamma_0\) and \(\gamma_1\) in a domain \(D\) are homotopic in \(D\) if one can be deformed continuously into the other through closed curves lying in \(D\). The homotopy form of Cauchy's theorem says that if \(f\) is analytic on \(D\) and \(\gamma_0\), \(\gamma_1\) are homotopic in \(D\), then \(\oint_{\gamma_0} f\,dz = \oint_{\gamma_1} f\,dz\). A domain is simply connected exactly when every closed curve is homotopic to a constant curve, whose integral is 0, so the basic theorem is a special case. On the punctured plane the unit circle is not homotopic to a point, because any deformation would have to pass through the origin; the value \(2\pi i\) of the integral of \(\frac{1}{z}\) is a certificate of that topological fact. Integration is thus a tool for detecting holes, an idea that grew into the de Rham cohomology of modern geometry.

Deformation of contours

The most useful form of Cauchy's theorem in practice concerns regions with holes. Let \(\gamma_1\) and \(\gamma_2\) be positively oriented simple closed contours with \(\gamma_2\) inside \(\gamma_1\), and suppose \(f\) is analytic on an open set containing the closed ring-shaped region between them. Then

\[\oint_{\gamma_1} f\,dz = \oint_{\gamma_2} f\,dz.\]

Proof sketch. Join \(\gamma_1\) to \(\gamma_2\) by two short cross-cuts, dividing the region between them into two pieces, each bounded by a closed contour on which \(f\) is analytic inside. Traverse each piece anticlockwise. By Cauchy's theorem each boundary integral is 0. Adding them, the cross-cuts are traversed once in each direction and cancel, the outer contour \(\gamma_1\) appears anticlockwise and the inner \(\gamma_2\) clockwise. So \(\oint_{\gamma_1} f\,dz - \oint_{\gamma_2} f\,dz = 0\). \(\square\)

In words: a contour may be deformed freely without changing the integral, as long as it does not pass over a point where the function fails to be analytic. Consequently, for any positively oriented simple closed contour \(\gamma\) round \(z_0\),

\[\oint_\gamma\frac{dz}{z - z_0} = 2\pi i,\]

since \(\gamma\) can be shrunk to a small circle about \(z_0\), where lesson 6 gives \(2\pi i\). The same argument with several inner contours shows that if \(f\) is analytic inside \(\gamma\) except at points \(z_1, \dots, z_n\), then \(\oint_\gamma f\,dz = \sum_k \oint_{C_k} f\,dz\), where \(C_k\) are small circles round the \(z_k\). This is the skeleton of the residue theorem.

import cmath, math

def integrate(f, gamma, dgamma, n=4000):
    """Simpson's rule on [0, 2*pi] for a closed parametrized contour."""
    h = 2*math.pi / n
    s = 0
    for k in range(n + 1):
        t = k*h
        w = 1 if k in (0, n) else (4 if k % 2 else 2)
        s += w * f(gamma(t)) * dgamma(t)
    return s * h / 3

# Three different closed contours round the origin.
contours = {
    "unit circle": (lambda t: cmath.exp(1j*t), lambda t: 1j*cmath.exp(1j*t)),
    "ellipse 3x1": (lambda t: 3*math.cos(t) + 1j*math.sin(t), lambda t: -3*math.sin(t) + 1j*math.cos(t)),
    "off-centre":  (lambda t: 0.5 + 2*cmath.exp(1j*t), lambda t: 2j*cmath.exp(1j*t)),
}
funcs = {"z^2 e^z (entire)": lambda z: z*z*cmath.exp(z),
         "cos z / (z - 5)":  lambda z: cmath.cos(z)/(z - 5),
         "1/z":              lambda z: 1/z,
         "e^z / z":          lambda z: cmath.exp(z)/z}
for fname, f in funcs.items():
    vals = [integrate(f, g, dg) for g, dg in contours.values()]
    print(f"{fname:18s}", "  ".join(f"{v.real:+.8f}{v.imag:+.8f}i" for v in vals))
print("2*pi*i =", 2j*math.pi)

# Partial fractions example: circle |z - 1| = 1 round 1 only.
v = integrate(lambda z: 1/(z*z - 1), lambda t: 1 + cmath.exp(1j*t), lambda t: 1j*cmath.exp(1j*t))
print("integral of 1/(z^2-1) round |z-1|=1:", v, " pi*i =", math.pi*1j)

z^2 e^z (entire)   +0.00000000+0.00000000i  +0.00000000+0.00000000i  +0.00000000+0.00000000i
cos z / (z - 5)    +0.00000000+0.00000000i  +0.00000000+0.00000000i  +0.00000000+0.00000000i
1/z                +0.00000000+6.28318531i  -0.00000000+6.28318531i  -0.00000000+6.28318531i
e^z / z            +0.00000000+6.28318531i  +0.00000000+6.28318531i  -0.00000000+6.28318531i
2*pi*i = 6.283185307179586j
integral of 1/(z^2-1) round |z-1|=1: (-4.642628913757749e-17+3.1415926535897944j)  pi*i = 3.141592653589793j

Common mistakes