Cauchy's integral theorem

In mathematics, the Cauchy integral theorem (also known as the Cauchy–Goursat theorem) in complex analysis, named after Augustin-Louis Cauchy (and Édouard Goursat), is an important statement about line integrals for holomorphic functions in the complex plane. Essentially, it says that if $$f(z)$$ is holomorphic in a simply connected domain Ω, then for any simply closed contour $$C$$ in Ω, that contour integral is zero.

$$\int_C f(z)\,dz = 0. $$

Fundamental theorem for complex line integrals
If $f(z)$ is a holomorphic function on an open region $U$, and $$\gamma$$ is a curve in $U$ from $$z_0$$ to $$z_1$$ then, $$\int_{\gamma}f'(z) \, dz = f(z_1)-f(z_0).$$

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

Formulation on simply connected regions
Let $$U \subseteq \Complex$$ be a simply connected open set, and let $$f: U \to \Complex$$ be a holomorphic function. Let $$\gamma: [a,b] \to U$$ be a smooth closed curve. Then: $$\int_\gamma f(z)\,dz = 0. $$ (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 $$U \subseteq \Complex$$ be an open set, and let $$f: U \to \Complex$$ be a holomorphic function. Let $$\gamma: [a,b] \to U$$ be a smooth closed curve. If $$\gamma$$ is homotopic to a constant curve, then: $$\int_\gamma f(z)\,dz = 0. $$ (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 $$\gamma$$ does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve: $$\gamma(t) = e^{it} \quad t \in \left[0, 2\pi\right] ,$$ which traces out the unit circle. Here the following integral: $$\int_{\gamma} \frac{1}{z}\,dz = 2\pi i \neq 0, $$ is nonzero. The Cauchy integral theorem does not apply here since $$f(z) = 1/z$$ is not defined at $$z = 0$$. Intuitively, $$\gamma$$ surrounds a "hole" in the domain of $$f$$, so $$\gamma$$ cannot be shrunk to a point without exiting the space. Thus, the theorem does not apply.

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 $$U_{z_0} = \{ z : \left|z-z_{0}\right| < r\}$$, for $$z_0 \in \Complex$$, qualifies. The condition is crucial; consider $$\gamma(t) = e^{it} \quad t \in \left[0, 2\pi\right]$$ which traces out the unit circle, and then the path integral $$\oint_\gamma \frac{1}{z}\,dz = \int_0^{2\pi} \frac{1}{e^{it}}(ie^{it} \,dt) = \int_0^{2\pi}i\,dt = 2\pi i $$ 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 $$\Complex$$, let $$f: U \to \Complex$$ be a holomorphic function, and let $$\gamma$$ 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 $$\int_\gamma f(z)\,dz=F(b)-F(a).$$

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

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 $$f=u+iv$$ must satisfy the Cauchy–Riemann equations in the region bounded by $\gamma$, 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 $$dz$$ into their real and imaginary components:

$$ f=u+iv $$ $$ dz=dx+i\,dy $$

In this case we have $$\oint_\gamma f(z)\,dz = \oint_\gamma (u+iv)(dx+i\,dy) = \oint_\gamma (u\,dx-v\,dy) +i\oint_\gamma (v\,dx+u\,dy)$$

By Green's theorem, we may then replace the integrals around the closed contour $$\gamma$$ with an area integral throughout the domain $$D$$ that is enclosed by $$\gamma$$ as follows:

$$\oint_\gamma (u\,dx-v\,dy) = \iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right) \,dx\,dy $$ $$\oint_\gamma (v\,dx+u\,dy) = \iint_D \left( \frac{\partial u}{\partial x} -\frac{\partial v}{\partial y} \right) \,dx\,dy $$

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

$$\iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right )\,dx\,dy = \iint_D \left( \frac{\partial u}{\partial y} - \frac{\partial u}{\partial y} \right ) \, dx \, dy =0$$ $$\iint_D \left( \frac{\partial u}{\partial x}-\frac{\partial v}{\partial y} \right )\,dx\,dy = \iint_D \left( \frac{\partial u}{\partial x} - \frac{\partial u}{\partial x} \right ) \, dx \, dy = 0$$

This gives the desired result $$\oint_\gamma f(z)\,dz = 0$$