Cauchy-Goursat Theorem
Cauchy-Goursat is the theorem concerning the integral of a holomorphic function over a closed contour. A simple closed contour is designated as positively oriented if, upon traversing the curve, its interior region is consistently to the left, for example, a circle traversed in the counterclockwise direction.
The theorem, first established by Cauchy under the assumption of a continuous derivative and later generalized by Goursat to only require holomorphicity, can be stated for a triangular path as follows.
Theorem: let \mathcal{D} be an open subset of the complex plane \mathbb{C}, and let T be a triangular contour within \mathcal{D} such that the interior of T is also entirely contained in \mathcal{D}.
If a function f: \mathcal{D} \to \mathbb{C} is holomorphic in \mathcal{D}, then the contour integral of f along T is zero.
\oint_T f(z) \mathrm{d}z = 0
The proof proceeds by constructing a sequence of nested triangles and exploiting the definition of holomorphicity.
Proof: let T^{(0)} be the original triangle T. We subdivide T^{(0)} by bisecting its three sides. Connecting these midpoints forms four smaller, congruent triangles, which we denote by T^{(0)}_j for j=1, 2, 3, 4.
We assign an orientation to each T^{(0)}_j consistent with the positive orientation of T^{(0)}. This geometric construction ensures that the integrals along the interior segments, each traversed in opposite directions, cancel each other out.
Consequently, the integral over the original triangle is the sum of the integrals over the four smaller triangles:
\oint_{T^{(0)}} f(z) \mathrm{d}z = \sum_{j=1}^4 \oint_{T^{(0)}_j} f(z) \mathrm{d}z
By applying the triangle inequality to this sum, we obtain a bound on the magnitude of the total integral:
\left|\oint_{T^{(0)}} f(z) \mathrm{d}z\right| \le \sum_{j=1}^4 \left| \oint_{T^{(0)}_j} f(z) \mathrm{d}z\right|
From this inequality, it follows that for at least one of the smaller triangles, say T^{(1)}, the magnitude of the integral must satisfy the following relation.
\left|\oint_{T^{(1)}} f(z) \mathrm{d}z\right| \ge \frac{1}{4} \left|\oint_{T^{(0)}} f(z) \mathrm{d}z\right|
We select such a triangle T^{(1)} and repeat this subdivision process inductively.
After n iterations of this procedure, we obtain a sequence of triangles T^{(n)} where each T^{(n)} is selected from the subdivision of T^{(n-1)} to have the maximal integral magnitude.
This yields the general inequality:
\left|\oint_T f(z) \mathrm{d}z\right| \le 4^n \left| \oint_{T^{(n)}} f(z) \mathrm{d}z\right|
Let \triangle^{(n)} be the closed triangular region bounded by the contour T^{(n)}. This construction generates a sequence of nested compact sets \dots \subset \triangle^{(n)} \subset \dots \subset \triangle^{(1)} \subset \triangle^{(0)}.
By Cantor’s intersection theorem, there exists a unique point z_0 common to all these regions:
\{z_0\} = \bigcap_{n=0}^\infty \triangle^{(n)}
Since f is holomorphic throughout \mathcal{D} and z_0 \in \mathcal{D}, the function is differentiable at z_0. The definition of the complex derivative allows us to write a first-order approximation of f(z) for z near z_0:
f(z) = f(z_0) + f^\prime(z_0)(z - z_0) + \psi(z)(z-z_0)
The function \psi(z) captures the remainder term of the linear approximation and satisfies \lim_{z \to z_0} \psi(z) = 0.
We now evaluate the integral of f(z) over the contour T^{(n)}.
\oint_{T^{(n)}} f(z) \mathrm{d}z = \oint_{T^{(n)}} \left( f(z_0) + f^\prime(z_0)(z-z_0) \right) \mathrm{d}z + \oint_{T^{(n)}} \psi(z)(z-z_0) \mathrm{d}z
The integrand in the first term on the right-hand side, g(z) = f(z_0) + f^\prime(z_0)(z-z_0), is a polynomial in z.
It possesses an antiderivative:
G(z) = f(z_0)z + \frac{1}{2}f^\prime(z_0)(z-z_0)^2
which is defined over the entire complex plane. The fundamental theorem of complex integration asserts that the integral of a function with an antiderivative over any closed path is zero. Therefore, the first integral vanishes.
\oint_{T^{(n)}} \left( f(z_0) + f^\prime(z_0)(z-z_0) \right) \mathrm{d}z = 0
This simplifies the expression for the integral over T^{(n)} to just the remainder term.
\oint_{T^{(n)}} f(z) \mathrm{d}z = \oint_{T^{(n)}} \psi(z)(z-z_0) \mathrm{d}z
To bound this integral, we use the ML-inequality. Let p^{(n)} and d^{(n)} be the perimeter and diameter of the triangle T^{(n)}, respectively. For any point z on the contour T^{(n)}, the distance |z - z_0| is bounded by the diameter, |z - z_0| \le d^{(n)}.
As \lim_{z \to z_0} \psi(z) = 0, for any given \varepsilon > 0, there exists a sufficiently large integer N such that for all n \ge N, the triangle \triangle^{(n)} is contained within a neighborhood of z_0 where |\psi(z)| < \varepsilon.
This gives us a bound for the magnitude of the integral:
\left| \oint_{T^{(n)}} f(z) \mathrm{d}z \right| = \left| \oint_{T^{(n)}} \psi(z)(z-z_0) \mathrm{d}z \right| \le \left( \sup_{z \in T^{(n)}} |\psi(z)(z-z_0)| \right) p^{(n)} \le (\varepsilon d^{(n)}) p^{(n)}
Substituting this result into our main inequality, we find:
\left|\oint_T f(z) \mathrm{d}z\right| \le 4^n |\oint_{T^{(n)}} f(z) \mathrm{d}z| \le 4^n \varepsilon d^{(n)} p^{(n)}
Using the geometric properties of our construction, detailed in the subsequent lemma, we have d^{(n)} = d^{(0)}/2^n and p^{(n)} = p^{(0)}/2^n. The inequality becomes:
\left|\oint_T f(z) \mathrm{d}z\right| \le 4^n \varepsilon \left(\frac{d^{(0)}}{2^n}\right) \left(\frac{p^{(0)}}{2^n}\right) = 4^n \varepsilon \frac{d^{(0)}p^{(0)}}{4^n} = \varepsilon d^{(0)}p^{(0)}
Since the perimeter p^{(0)} and diameter d^{(0)} of the original triangle T are fixed values, and \varepsilon can be made arbitrarily small, the magnitude of the integral must be less than any positive number. The only non-negative value with this property is zero:
\left|\oint_T f(z) \mathrm{d}z\right| = 0
This implies the desired result:
\oint_T f(z) \mathrm{d}z = 0
The proof relies on the scaling of the perimeter and diameter of the nested triangles.
Lemma: let p^{(n)} and d^{(n)} denote the perimeter and diameter of the triangle T^{(n)}, respectively. Then for n \ge 0, the following relations hold.
\begin{aligned} p^{(n)} & = \frac{p^{(0)}}{2^n} \\ d^{(n)} & = \frac{d^{(0)}}{2^n} \end{aligned}
Proof: the triangle T^{(n)} is constructed from T^{(n-1)} through bisection of its sides. This process creates four smaller triangles that are congruent to each other and similar to the parent triangle T^{(n-1)}. The side lengths of any of these smaller triangles are precisely half the corresponding side lengths of T^{(n-1)}.
Since the perimeter of a triangle is the sum of its side lengths, it scales linearly with the side lengths. Therefore, p^{(n)} = \frac{1}{2}p^{(n-1)}.
The diameter of a convex set such as a triangle is the greatest distance between any two of its points, which corresponds to the length of its longest side. This also scales linearly, giving d^{(n)} = \frac{1}{2}d^{(n-1)}.
Applying these relations recursively from n down to 0 establishes the formulas:
\begin{aligned} p^{(n)} & = \frac{1}{2} p^{(n-1)} = \left(\frac{1}{2}\right)^2 p^{(n-2)} = \dots = \left(\frac{1}{2}\right)^n p^{(0)} \\ d^{(n)} & = \frac{1}{2} d^{(n-1)} = \left(\frac{1}{2}\right)^2 d^{(n-2)} = \dots = \left(\frac{1}{2}\right)^n d^{(0)} \end{aligned}
This completes the demonstration.
Proof with Green’s theorem
The proof presented by Goursat requires only that the function f(z) be holomorphic.
Historically, a proof was first given by Cauchy under the more restrictive condition that the partial derivatives of the components of f(z) are continuous.
This stronger assumption allows for a direct application of Green’s theorem from vector calculus, offering an alternative and more immediate derivation of the result. We present the proof below.
Proof: let us assume that f(z) is holomorphic in an open set \mathcal{D} and that its derivative f'(z) is continuous in \mathcal{D}.
This implies that if we express f(z) in terms of its real and imaginary parts:
f(x+iy) = u(x,y) + i v(x,y)
the partial derivatives of u and v with respect to x and y are continuous.
We begin by decomposing the complex contour integral into its real and imaginary components. The function f(z) and the differential element \mathrm{d}z are expressed as:
\begin{aligned} f(z) &= u(x,y) + i v(x,y) \\ \mathrm{d}z &= \mathrm{d}x + i \mathrm{d}y \end{aligned}
Substituting these into the integral over a simple closed contour T enclosing a region \mathcal{R} yields:
\oint_T f(z) \mathrm{d}z = \oint_T (u+iv)(\mathrm{d}x + i \mathrm{d}y)
Expanding the product in the integrand gives:
\oint_T (u \mathrm{d}x + i u \mathrm{d}y + i v \mathrm{d}x - v \mathrm{d}y)
By separating the real and imaginary parts of the integral, we obtain two real line integrals.
\oint_T f(z) \mathrm{d}z = \oint_T (u \mathrm{d}x - v \mathrm{d}y) + i \oint_T (v \mathrm{d}x + u \mathrm{d}y)
At this point, we can apply Green’s theorem.
Green’s theorem states that for a region \mathcal{R} bounded by a positively oriented, piecewise smooth, simple closed curve T, and for functions P(x,y) and Q(x,y) with continuous partial derivatives in an open region containing \mathcal{R}, the following holds:
\oint_T (P \mathrm{d}x + Q \mathrm{d}y) = \iint_{\mathcal{R}} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathrm{d}x \mathrm{d}y
Applying this theorem to the real part of our complex integral, with P=u and Q=-v, we find:
\oint_T (u \mathrm{d}x - v \mathrm{d}y) = \iint_{\mathcal{R}} \left( -\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) \mathrm{d}x \mathrm{d}y
Similarly, for the imaginary part, with P=v and Q=u, we have:
\oint_T (v \mathrm{d}x + u \mathrm{d}y) = \iint_{\mathcal{R}} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) \mathrm{d}x \mathrm{d}y
The hypothesis that f(z) is holomorphic requires its real and imaginary parts, u and v, to satisfy the Cauchy-Riemann equations throughout the domain \mathcal{D}, and therefore within \mathcal{R}:
\begin{aligned} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} \end{aligned}
We now substitute these conditions into the double integrals. For the first integral, the integrand becomes:
-\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = - \left(-\frac{\partial u}{\partial y}\right) - \frac{\partial u}{\partial y} = \frac{\partial u}{\partial y} - \frac{\partial u}{\partial y} = 0
For the second integral, the integrand is:
\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} = \frac{\partial v}{\partial y} - \frac{\partial v}{\partial y} = 0
Since both integrands are identically zero over the entire region \mathcal{R}, the double integrals evaluate to zero.
\begin{aligned} & \iint_{\mathcal{R}} \left( -\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) \mathrm{d}x \mathrm{d}y = 0 \\ & \iint_{\mathcal{R}} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) \mathrm{d}x \mathrm{d}y = 0 \end{aligned}
This leads to the conclusion that the original complex integral is zero.
\oint_T f(z) \mathrm{d}z = 0 + i \cdot 0 = 0
This completes the proof under the stronger assumption of the continuity of f'(z).
Goursat’s proof has the ability to establish this same result without recourse to this condition, relying solely on the existence of the derivative f'(z) at every point in the domain.
This distinction shows that holomorphicity alone is a sufficient condition to guarantee the vanishing of the integral, a fact that is not immediately evident from the perspective of vector calculus.