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Mathematical Methods

The complex integral, Cauchy's theorem and Cauchy's integral formula

Integrals along a curve, proof of Cauchy's theorem with Green, deforming paths, the integral formula, derivatives of every order, Liouville and the fundamental theorem of algebra.

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The integral along a curve in the complex plane is the centre of the whole theory. Cauchy’s theorem says that for holomorphic functions the integral around a closed curve is zero, and Cauchy’s integral formula says that the values of a function inside a curve are determined by its values on the curve. From here follow results with no real counterpart, such as the fact that a function differentiable once is differentiable infinitely many times.

The complex integral

Let \gamma be a regular curve parametrised by z(t), t\in[a, b]. The integral of f along \gamma is

    \[ \int_\gamma f(z)\,dz = \int_a^b f(z(t))\,z'(t)\,dt \]

Writing f = u + iv and dz = dx + i\,dy shows it is a combination of Analysis 2 line integrals:

    \[ \int_\gamma f\,dz = \int_\gamma(u\,dx - v\,dy) + i\int_\gamma(v\,dx + u\,dy) \]

The fundamental example. Let \gamma be the circle with centre z_0 and radius r, traversed once anticlockwise: z = z_0 + re^{it}, dz = ire^{it}dt. Then

    \[ \oint_\gamma\frac{dz}{z - z_0} = \int_0^{2\pi}\frac{ire^{it}}{re^{it}}\,dt = 2\pi i \]

and, for every integer n \ne -1, \oint_\gamma(z - z_0)^n\,dz = \int_0^{2\pi}ir^{n+1}e^{i(n+1)t}\,dt = 0. These two integrals are the key to this lesson and the next.

A useful bound. \left|\int_\gamma f\,dz\right| \le M\cdot L, where M is the maximum of |f| on the curve and L its length.

Cauchy’s theorem

Theorem. If f is holomorphic in a simply connected domain D, then for every closed curve \gamma in D

    \[ \oint_\gamma f(z)\,dz = 0 \]

Proof (assuming f' is continuous)

Apply Green’s theorem from Analysis 2 to the two line integrals, with R the region enclosed by \gamma:

    \[ \oint(u\,dx - v\,dy) = \iint_R\left(-\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right)dx\,dy,\qquad \oint(v\,dx + u\,dy) = \iint_R\left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\right)dx\,dy \]

By the Cauchy-Riemann equations both integrands are zero. Goursat later showed that the continuity of f' is not needed.

Consequences.

  • The integral between two points does not depend on the path, and f has an antiderivative F with F' = f.
  • A curve can be deformed freely without changing the integral, as long as it does not cross points where f is not holomorphic. In the figure, the integral along \gamma equals the sum of the integrals round the small circles about the inner singularities z_1 and z_2; the outer singularity z_3 does not count.
A closed curve in the complex plane enclosing two singularities and leaving a third outside
The curve \gamma, traversed anticlockwise, encloses the singularities z_1 and z_2 but not z_3. By Cauchy’s theorem, the integral along \gamma can be replaced by those round the small dashed circles about z_1 and z_2.

Why holomorphy matters. With f(z) = 1/z and the unit circle the integral is 2\pi i, not zero: the function is not holomorphic at the origin, which lies inside the curve. It is the same phenomenon as the field with a “hole” seen in Analysis 2 for conservative fields.

Cauchy’s integral formula

Theorem. If f is holomorphic inside and on a simple closed curve \gamma traversed anticlockwise, and z_0 is inside \gamma, then

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

Proof

The function \frac{f(z)}{z - z_0} is holomorphic except at z_0, so \gamma can be shrunk to a tiny circle of radius r about z_0. Write

    \[ \oint\frac{f(z)}{z - z_0}dz = f(z_0)\oint\frac{dz}{z - z_0} + \oint\frac{f(z) - f(z_0)}{z - z_0}dz \]

The first term is 2\pi i\,f(z_0). In the second, the numerator tends to zero by continuity of f, and the M\cdot L bound gives at most \frac{\max|f(z) - f(z_0)|}{r}\cdot 2\pi r, which tends to zero with r. Since the integral does not depend on r, the second term is exactly zero.

The meaning. The values of f inside the curve are determined by its values on the curve. For a circle centred at z_0, the formula says that f(z_0) is the average of the values on the circle (the mean value property): the same fact by which the steady temperature at the centre of a disc is the average of the temperature on its edge.

Consequences

Derivatives of every order. Differentiating the formula under the integral sign with respect to z_0:

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

A holomorphic function has derivatives of every order, all holomorphic. In the real case this is not so at all: x|x| has a first derivative but no second.

Liouville’s theorem. A function holomorphic on the whole of \mathbb C and bounded is constant. With the formula for f' on a circle of radius R, |f'(z_0)| \le \frac{M}{R}, which tends to zero as R\to\infty. It is also why complex sine and cosine cannot be bounded.

The fundamental theorem of algebra. Every non-constant polynomial p(z) has at least one complex root. If it had none, 1/p(z) would be holomorphic on the whole plane and bounded (because |p(z)|\to\infty as |z|\to\infty), hence constant by Liouville: a contradiction. It follows that a polynomial of degree n has exactly n complex roots, counted with multiplicity. It is the result that guarantees that the characteristic polynomial of every linear circuit or system always has its poles.

Worked example. \oint_{|z| = 2}\frac{e^z}{z - 1}dz: e^z is holomorphic everywhere and z_0 = 1 is inside the circle, so the integral is 2\pi i\,e.

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