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 ${\displaystyle f(z)}${\displaystyle f(z)} is holomorphic in a simply connected domain Ω, then for any simply closed contour ${\displaystyle C}${\displaystyle C} in Ω, that contour integral is zero.

$${\displaystyle \int _{C}f(z),円dz=0.}$${\displaystyle \int _{C}f(z),円dz=0.}

If f(z) is a holomorphic function on an open region U, and ${\displaystyle \gamma }${\displaystyle \gamma } is a curve in U from ${\displaystyle z_{0}}${\displaystyle z_{0}} to ${\displaystyle z_{1}}${\displaystyle z_{1}} then, $${\displaystyle \int _{\gamma }f'(z),円dz=f(z_{1})-f(z_{0}).}$${\displaystyle \int _{\gamma }f'(z),円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 ${\textstyle \int _{\gamma }f(z),円dz}${\textstyle \int _{\gamma }f(z),円dz} is path independent for all paths in U.

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

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

(Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy (within ${\displaystyle U}${\displaystyle 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.

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

As Édouard Goursat showed, Cauchy's integral theorem can be proven assuming only that the complex derivative ${\displaystyle f'(z)}${\displaystyle f'(z)} exists everywhere in ${\displaystyle U}${\displaystyle 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 ${\displaystyle U}${\displaystyle U} be simply connected means that ${\displaystyle U}${\displaystyle U} has no "holes" or, in homotopy terms, that the fundamental group of ${\displaystyle U}${\displaystyle U} is trivial; for instance, every open disk ${\displaystyle U_{z_{0}}=\{z:\left|z-z_{0}\right|<r\}}${\displaystyle U_{z_{0}}=\{z:\left|z-z_{0}\right|<r\}}, for ${\displaystyle z_{0}\in \mathbb {C} }${\displaystyle z_{0}\in \mathbb {C} }, qualifies. The condition is crucial; consider $${\displaystyle \gamma (t)=e^{it}\quad t\in \left[0,2\pi \right]}$${\displaystyle \gamma (t)=e^{it}\quad t\in \left[0,2\pi \right]} which traces out the unit circle, and then the path integral $${\displaystyle \oint _{\gamma }{\frac {1}{z}},円dz=\int _{0}^{2\pi }{\frac {1}{e^{it}}}(ie^{it},円dt)=\int _{0}^{2\pi }i,円dt=2\pi i}$${\displaystyle \oint _{\gamma }{\frac {1}{z}},円dz=\int _{0}^{2\pi }{\frac {1}{e^{it}}}(ie^{it},円dt)=\int _{0}^{2\pi }i,円dt=2\pi i} is nonzero; the Cauchy integral theorem does not apply here since ${\displaystyle f(z)=1/z}${\displaystyle f(z)=1/z} is not defined (and is certainly not holomorphic) at ${\displaystyle z=0}${\displaystyle 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 ${\displaystyle U}${\displaystyle U} be a simply connected open subset of ${\displaystyle \mathbb {C} }${\displaystyle \mathbb {C} }, let ${\displaystyle f:U\to \mathbb {C} }${\displaystyle f:U\to \mathbb {C} } be a holomorphic function, and let ${\displaystyle \gamma }${\displaystyle \gamma } be a piecewise continuously differentiable path in ${\displaystyle U}${\displaystyle U} with start point ${\displaystyle a}${\displaystyle a} and end point ${\displaystyle b}${\displaystyle b}. If ${\displaystyle F}${\displaystyle F} is a complex antiderivative of ${\displaystyle f}${\displaystyle f}, then $${\displaystyle \int _{\gamma }f(z),円dz=F(b)-F(a).}$${\displaystyle \int _{\gamma }f(z),円dz=F(b)-F(a).}

The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given ${\displaystyle U}${\displaystyle U}, a simply connected open subset of ${\displaystyle \mathbb {C} }${\displaystyle \mathbb {C} }, we can weaken the assumptions to ${\displaystyle f}${\displaystyle f} being holomorphic on ${\displaystyle U}${\displaystyle U} and continuous on ${\textstyle {\overline {U}}}${\textstyle {\overline {U}}} and ${\displaystyle \gamma }${\displaystyle \gamma } a rectifiable simple loop in ${\textstyle {\overline {U}}}${\textstyle {\overline {U}}}.^[1]

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

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

$${\displaystyle f=u+iv}$${\displaystyle f=u+iv} $${\displaystyle dz=dx+i,円dy}$${\displaystyle dz=dx+i,円dy}

In this case we have $${\displaystyle \oint _{\gamma }f(z),円dz=\oint _{\gamma }(u+iv)(dx+i,円dy)=\oint _{\gamma }(u,円dx-v,円dy)+i\oint _{\gamma }(v,円dx+u,円dy)}$${\displaystyle \oint _{\gamma }f(z),円dz=\oint _{\gamma }(u+iv)(dx+i,円dy)=\oint _{\gamma }(u,円dx-v,円dy)+i\oint _{\gamma }(v,円dx+u,円dy)}

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

$${\displaystyle \oint _{\gamma }(u,円dx-v,円dy)=\iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right),円dx,円dy}$${\displaystyle \oint _{\gamma }(u,円dx-v,円dy)=\iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right),円dx,円dy} $${\displaystyle \oint _{\gamma }(v,円dx+u,円dy)=\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right),円dx,円dy}$${\displaystyle \oint _{\gamma }(v,円dx+u,円dy)=\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right),円dx,円dy}

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

We therefore find that both integrands (and hence their integrals) are zero

$${\displaystyle \iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right),円dx,円dy=\iint _{D}\left({\frac {\partial u}{\partial y}}-{\frac {\partial u}{\partial y}}\right),円dx,円dy=0}$${\displaystyle \iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right),円dx,円dy=\iint _{D}\left({\frac {\partial u}{\partial y}}-{\frac {\partial u}{\partial y}}\right),円dx,円dy=0} $${\displaystyle \iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right),円dx,円dy=\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial u}{\partial x}}\right),円dx,円dy=0}$${\displaystyle \iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right),円dx,円dy=\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial u}{\partial x}}\right),円dx,円dy=0}

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

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

• Kodaira, Kunihiko (2007), Complex Analysis, Cambridge Stud. Adv. Math., 107, CUP, 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

• "Cauchy integral theorem", Encyclopedia of Mathematics , EMS Press, 2001 [1994]
• Weisstein, Eric W. "Cauchy Integral Theorem". MathWorld .
• Jeremy Orloff, 18.04 Complex Variables with Applications Spring 2018 Massachusetts Institute of Technology: MIT OpenCourseWare Creative Commons.

• Augustin-Louis Cauchy
• Eponymous theorems of mathematics
• Theorems in complex analysis

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