Cauchy's integral theorem

Statement

Fundamental theorem for complex line integrals

If ⁠ $f (z)$ ⁠ is a holomorphic function on an open region ⁠ $U$ ⁠, and $γ$ is a curve in ⁠ $U$ ⁠ from $z_{0}$ to $z_{1}$ then, $\int_{γ} f^{'} (z) d z = f (z_{1}) - f (z_{0}) .$

Also, when ⁠ $f (z)$ ⁠ has a single-valued antiderivative in an open region ⁠ $U$ ⁠, then the path integral ⁠ $\int_{γ} f (z) d z$ ⁠ is path independent for all paths in ⁠ $U$ ⁠.

Formulation on simply connected regions

Let $U \subseteq C$ be a simply connected open set, and let $f : U \to C$ be a holomorphic function. Let $γ : [a, b] \to U$ be a smooth closed curve. Then: $\int_{γ} f (z) d z = 0.$ (The condition that $U$ be simply connected means that $U$ has no "holes", or in other words, that the fundamental group of $U$ is trivial.)

General formulation

Let $U \subseteq C$ be an open set, and let $f : U \to C$ be a holomorphic function. Let $γ : [a, b] \to U$ be a smooth closed curve. If $γ$ is homotopic to a constant curve, then: $\int_{γ} f (z) d z = 0,$ where ⁠ $z \in U$ ⁠.

(Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy (within $U$ ) from the curve to the constant curve. Intuitively, this means that one can shrink the curve into a point without exiting the space.) The first version is a special case of this because on a simply connected set, every closed curve is homotopic to a constant curve.

Main example

In both cases, it is important to remember that the curve $γ$ does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve: $γ (t) = e^{i t} t \in [0, 2 π],$ which traces out the unit circle. Here the following integral: $\int_{γ} \frac{1}{z} d z = 2 π i \neq 0,$ is nonzero. The Cauchy integral theorem does not apply here since $f (z) = 1 / z$ is not defined at ⁠ $z = 0$ ⁠. Intuitively, $γ$ surrounds a "hole" in the domain of ⁠ $f$ ⁠, so $γ$ cannot be shrunk to a point without exiting the space. Thus, the theorem does not apply.

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Discussion

As Édouard Goursat showed, Cauchy's integral theorem can be proven assuming only that the complex derivative $f^{'} (z)$ exists everywhere in ⁠ $U$ ⁠. This is significant because one can then prove Cauchy's integral formula for these functions, and from that deduce these functions are infinitely differentiable.

The condition that $U$ be simply connected means that $U$ has no "holes" or, in homotopy terms, that the fundamental group of $U$ is trivial; for instance, every open disk ⁠ $U_{z_{0}} = {z : | z - z_{0} | < r}$ ⁠, for ⁠ $z_{0} \in C$ ⁠, qualifies. The condition is crucial; consider $γ (t) = e^{i t} t \in [0, 2 π]$ which traces out the unit circle, and then the path integral $\oint_{γ} \frac{1}{z} d z = \int_{0}^{2 π} \frac{1}{e^{i t}} (i e^{i t} d t) = \int_{0}^{2 π} i d t = 2 π i$ is nonzero; the Cauchy integral theorem does not apply here since $f (z) = 1 / z$ is not defined (and is certainly not holomorphic) at ⁠ $z = 0$ ⁠.

One important consequence of the theorem is that path integrals of holomorphic functions on simply connected domains can be computed in a manner familiar from the fundamental theorem of calculus: let $U$ be a simply connected open subset of ⁠ $C$ ⁠, let $f : U \to C$ be a holomorphic function, and let $γ$ be a piecewise continuously differentiable path in $U$ with start point $a$ and end point ⁠ $b$ ⁠. If $F$ is a complex antiderivative of ⁠ $f$ ⁠, then $\int_{γ} f (z) d z = F (b) - F (a) .$

The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given ⁠ $U$ ⁠, a simply connected open subset of ⁠ $C$ ⁠, we can weaken the assumptions to $f$ being holomorphic on $U$ and continuous on $\bar{U}$ and $γ$ a rectifiable simple loop in ⁠ $\bar{U}$ ⁠.

The Cauchy integral theorem leads to Cauchy's integral formula and the residue theorem.

Proof

If one assumes that the partial derivatives of a holomorphic function are continuous, the Cauchy integral theorem can be proven as a direct consequence of Green's theorem and the fact that the real and imaginary parts of $f = u + i v$ must satisfy the Cauchy–Riemann equations in the region bounded by ⁠ $γ$ ⁠, and moreover in the open neighborhood U of this region. Cauchy provided this proof, but it was later proven by Goursat without requiring techniques from vector calculus, or the continuity of partial derivatives.

We can break the integrand ⁠ $f$ ⁠, as well as the differential $d z$ into their real and imaginary components: $f = u + i v$ $d z = d x + i d y$

In this case we have $\oint_{γ} f (z) d z = \oint_{γ} (u + i v) (d x + i d y) = \oint_{γ} (u d x - v d y) + i \oint_{γ} (v d x + u d y)$

By Green's theorem, we may then replace the integrals around the closed contour $γ$ with an area integral throughout the domain $D$ that is enclosed by $γ$ as follows: $\oint_{γ} (u d x - v d y) = \iint_{D} (- \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}) d x d y$ $\oint_{γ} (v d x + u d y) = \iint_{D} (\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}) d x d y$

But as the real and imaginary parts of a function holomorphic in the domain ⁠ $D$ ⁠, $u$ and $v$ must satisfy the Cauchy–Riemann equations there: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ $\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$

We therefore find that both integrands (and hence their integrals) are zero: $\iint_{D} (- \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}) d x d y = \iint_{D} (\frac{\partial u}{\partial y} - \frac{\partial u}{\partial y}) d x d y = 0$ $\iint_{D} (\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}) d x d y = \iint_{D} (\frac{\partial u}{\partial x} - \frac{\partial u}{\partial x}) d x d y = 0$

This gives the desired result $\oint_{γ} f (z) d z = 0.$