Cauchy's integral theorem

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

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

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

Let ${\displaystyle U\subseteq \mathbb {C} }$  be an open set, and let ${\displaystyle f:U\to \mathbb {C} }$  be a holomorphic function. Let ${\displaystyle \gamma :[a,b]\to U}$  be a smooth closed curve. If ${\displaystyle \gamma }$  is homotopic to a constant curve, then: $${\displaystyle \int _{\gamma }f(z)\,dz=0.}$$  (Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy (within ${\displaystyle U}$ ) 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 ${\displaystyle \gamma }$  does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve: $${\displaystyle \gamma (t)=e^{it}\quad t\in \left[0,2\pi \right],}$$  which traces out the unit circle. Here the following integral: $${\displaystyle \int _{\gamma }{\frac {1}{z}}\,dz=2\pi i\neq 0,}$$  is nonzero. The Cauchy integral theorem does not apply here since ${\displaystyle f(z)=1/z}$  is not defined at ${\displaystyle z=0}$ . Intuitively, ${\displaystyle \gamma }$  surrounds a "hole" in the domain of ${\displaystyle f}$ , so ${\displaystyle \gamma }$  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 ${\displaystyle f'(z)}$  exists everywhere in ${\displaystyle U}$ . 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 ${\displaystyle U}$  be simply connected means that ${\displaystyle U}$  has no "holes" or, in homotopy terms, that the fundamental group of ${\displaystyle U}$  is trivial; for instance, every open disk ${\displaystyle U_{z_{0}}=\{z:\left|z-z_{0}\right|<r\}}$ , for ${\displaystyle z_{0}\in \mathbb {C} }$ , qualifies. The condition is crucial; consider $${\displaystyle \gamma (t)=e^{it}\quad t\in \left[0,2\pi \right]}$$  which traces out the unit circle, and then the path integral $${\displaystyle \oint _{\gamma }{\frac {1}{z}}\,dz=\int _{0}^{2\pi }{\frac {1}{e^{it}}}(ie^{it}\,dt)=\int _{0}^{2\pi }i\,dt=2\pi i}$$  is nonzero; the Cauchy integral theorem does not apply here since ${\displaystyle f(z)=1/z}$  is not defined (and is certainly not holomorphic) at ${\displaystyle z=0}$ .

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 ${\displaystyle U}$  be a simply connected open subset of ${\displaystyle \mathbb {C} }$ , let ${\displaystyle f:U\to \mathbb {C} }$  be a holomorphic function, and let ${\displaystyle \gamma }$  be a piecewise continuously differentiable path in ${\displaystyle U}$  with start point ${\displaystyle a}$  and end point ${\displaystyle b}$ . If ${\displaystyle F}$  is a complex antiderivative of ${\displaystyle f}$ , then $${\displaystyle \int _{\gamma }f(z)\,dz=F(b)-F(a).}$$ 

The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given ${\displaystyle U}$ , a simply connected open subset of ${\displaystyle \mathbb {C} }$ , we can weaken the assumptions to ${\displaystyle f}$  being holomorphic on ${\displaystyle U}$  and continuous on ${\textstyle {\overline {U}}}$  and ${\displaystyle \gamma }$  a rectifiable simple loop in ${\textstyle {\overline {U}}}$ .^[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 ${\displaystyle f=u+iv}$  must satisfy the Cauchy–Riemann equations in the region bounded by ${\displaystyle \gamma }$ , 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 ${\displaystyle f}$ , as well as the differential ${\displaystyle dz}$  into their real and imaginary components:

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

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

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

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

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

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

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

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

• 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.