Cauchy's integral theorem

Short description: Theorem in complex analysis

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.

$$\underset{C}{\int }f(z)dz=0.$$

Statement

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, $$\underset{\gamma }{\int }{f}^{\prime }(z)dz=f(z_1)-f(z_0).$$

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

Formulation on simply connected regions

Let $U\subseteq \mathbb{ℂ}$ be a simply connected open set, and let $f:U\to \mathbb{ℂ}$ be a holomorphic function. Let $\gamma :[a,b]\to U$ be a smooth closed curve. Then: $$\underset{\gamma }{\int }f(z)dz=0.$$ (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 $U\subseteq \mathbb{ℂ}$ be an open set, and let $f:U\to \mathbb{ℂ}$ be a holomorphic function. Let $\gamma :[a,b]\to U$ be a smooth closed curve. If $\gamma$ is homotopic to a constant curve, then: $$\underset{\gamma }{\int }f(z)dz=0.$$ (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: $$\gamma (t)=e^{it}t\in [0,2\pi ],$$ which traces out the unit circle. Here the following integral: $$\underset{\gamma }{\int }\frac{1}{z}dz=2\pi i\ne 0,$$ 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 $f^{\prime }(z)$ 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 $z_0\in \mathbb{ℂ}$, qualifies. The condition is crucial; consider $$\gamma (t)=e^{it}t\in [0,2\pi ]$$ which traces out the unit circle, and then the path integral $${\oint }_{\gamma }\frac{1}{z}dz=\underset{0}{\overset{2\pi }{\int }}\frac{1}{e^{it}}(i{e}^{it}dt)=\underset{0}{\overset{2\pi }{\int }}idt=2\pi i$$ 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 $\mathbb{ℂ}$, let $f:U\to \mathbb{ℂ}$ 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 $$\underset{\gamma }{\int }f(z)dz=F(b)-F(a).$$

The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given $U$, a simply connected open subset of $\mathbb{ℂ}$, we can weaken the assumptions to $f$ being holomorphic on $U$ and continuous on $\bar{U}$ and $\gamma$ 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 $f=u+iv$ 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:

$$f=u+iv$$ $$dz=dx+idy$$

In this case we have $${\oint }_{\gamma }f(z)dz={\oint }_{\gamma }(u+iv)(dx+idy)={\oint }_{\gamma }(udx-vdy)+i{\oint }_{\gamma }(vdx+udy)$$

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:

$${\oint }_{\gamma }(udx-vdy)=\underset{D}{\iint }(-\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y})dxdy$$ $${\oint }_{\gamma }(vdx+udy)=\underset{D}{\iint }(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y})dxdy$$

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

$$\underset{D}{\iint }(-\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y})dxdy=\underset{D}{\iint }(\frac{\partial u}{\partial y}-\frac{\partial u}{\partial y})dxdy=0$$ $$\underset{D}{\iint }(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y})dxdy=\underset{D}{\iint }(\frac{\partial u}{\partial x}-\frac{\partial u}{\partial x})dxdy=0$$

This gives the desired result $${\oint }_{\gamma }f(z)dz=0$$

See also

• Morera's theorem
• Methods of contour integration
• Star domain

References

• Kodaira, Kunihiko (2007), Complex Analysis, Cambridge Stud. Adv. Math., 107, Cambridge University Press, ISBN 978-0-521-80937-5 
• Ahlfors, Lars (2000), Complex Analysis, McGraw-Hill series in Mathematics, McGraw-Hill, ISBN 0-07-000657-1 
• Lang, Serge (2003), Complex Analysis, Springer Verlag GTM, Springer Verlag 
• Rudin, Walter (2000), Real and Complex Analysis, McGraw-Hill series in mathematics, McGraw-Hill 

External links

• Hazewinkel, Michiel, ed. (2001), "Cauchy integral theorem", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/c020900 
• Weisstein, Eric W.. "Cauchy Integral Theorem". http://mathworld.wolfram.com/CauchyIntegralTheorem.html. 
• Jeremy Orloff, 18.04 Complex Variables with Applications Spring 2018 Massachusetts Institute of Technology: MIT OpenCourseWare Creative Commons.