Cauchy's integral theorem

If {{math|f(z)}} is a holomorphic function on an open region {{mvar|U}}, and $\backslash gamma$ is a curve in {{mvar|U}} from $z\_0$ to $z\_1$ then,

$$\backslash int\_\{\backslash gamma\}f\text{'}(z)\; \backslash ,\; dz\; =\; f(z\_1)-f(z\_0).$$

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

Let $U\; \backslash subseteq\; \backslash Complex$ be a simply connected open set, and let $f:\; U\; \backslash to\; \backslash Complex$ be a holomorphic function. Let $\backslash gamma:\; [a,b]\; \backslash to\; U$ be a smooth closed curve. Then:

$$\backslash int\_\backslash gamma\; f(z)\backslash ,dz\; =\; 0.$$

(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.)

Let $U\; \backslash subseteq\; \backslash Complex$ be an open set, and let $f:\; U\; \backslash to\; \backslash Complex$ be a holomorphic function. Let $\backslash gamma:\; [a,b]\; \backslash to\; U$ be a smooth closed curve. If $\backslash gamma$ is homotopic to a constant curve, then:

$$\backslash int\_\backslash gamma\; f(z)\backslash ,dz\; =\; 0.$$where z є 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.

In both cases, it is important to remember that the curve $\backslash gamma$ does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve:

$$\backslash gamma(t)\; =\; e^\{it\}\; \backslash quad\; t\; \backslash in\; \backslash left[0,\; 2\backslash pi\backslash right]\; ,$$

which traces out the unit circle. Here the following integral:

$$\backslash int\_\{\backslash gamma\}\; \backslash frac\{1\}\{z\}\backslash ,dz\; =\; 2\backslash pi\; i\; \backslash neq\; 0\; ,$$

is nonzero. The Cauchy integral theorem does not apply here since $f(z)\; =\; 1/z$ is not defined at $z\; =\; 0$. Intuitively, $\backslash gamma$ surrounds a "hole" in the domain of $f$, so $\backslash gamma$ 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 $f\text{'}(z)$ 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 , for $z\_0\; \backslash in\; \backslash Complex$, qualifies. The condition is crucial; consider

$$\backslash gamma(t)\; =\; e^\{it\}\; \backslash quad\; t\; \backslash in\; \backslash left[0,\; 2\backslash pi\backslash right]$$

which traces out the unit circle, and then the path integral

$$\backslash oint\_\backslash gamma\; \backslash frac\{1\}\{z\}\backslash ,dz\; =\; \backslash int\_0^\{2\backslash pi\}\; \backslash frac\{1\}\{e^\{it\}\}(ie^\{it\}\; \backslash ,dt)\; =\; \backslash int\_0^\{2\backslash pi\}i\backslash ,dt\; =\; 2\backslash pi\; i$$

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 $\backslash Complex$, let $f:\; U\; \backslash to\; \backslash Complex$ be a holomorphic function, and let $\backslash gamma$ 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

$$\backslash int\_\backslash gamma\; f(z)\backslash ,dz=F(b)-F(a).$$

The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given $U$, a simply connected open subset of $\backslash Complex$, we can weaken the assumptions to $f$ being holomorphic on $U$ and continuous on $\backslash overline\{U\}$ and $\backslash gamma$ a rectifiable simple loop in $\backslash overline\{U\}$.{{Cite journal|last=Walsh|first=J. L.|date=1933-05-01|title=The Cauchy-Goursat Theorem for Rectifiable Jordan Curves|journal=Proceedings of the National Academy of Sciences|volume=19|issue=5|pages=540–541| doi=10.1073/pnas.19.5.540|pmid=16587781|pmc=1086062|issn=0027-8424|doi-access=free|bibcode=1933PNAS...19..540W }}

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

$$f=u+iv$$

$$dz=dx+i\backslash ,dy$$

In this case we have

$$\backslash oint\_\backslash gamma\; f(z)\backslash ,dz\; =\; \backslash oint\_\backslash gamma\; (u+iv)(dx+i\backslash ,dy)\; =\; \backslash oint\_\backslash gamma\; (u\backslash ,dx-v\backslash ,dy)\; +i\backslash oint\_\backslash gamma\; (v\backslash ,dx+u\backslash ,dy)$$

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

$$\backslash oint\_\backslash gamma\; (u\backslash ,dx-v\backslash ,dy)\; =\; \backslash iint\_D\; \backslash left(\; -\backslash frac\{\backslash partial\; v\}\{\backslash partial\; x\}\; -\backslash frac\{\backslash partial\; u\}\{\backslash partial\; y\}\; \backslash right)\; \backslash ,dx\backslash ,dy$$

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

But as the real and imaginary parts of a function holomorphic in the domain {{nowrap|$D$,}} $u$ and $v$ must satisfy the Cauchy–Riemann equations there:

$$\backslash frac\{\; \backslash partial\; u\; \}\{\; \backslash partial\; x\; \}\; =\; \backslash frac\{\; \backslash partial\; v\; \}\{\; \backslash partial\; y\; \}$$

$$\backslash frac\{\; \backslash partial\; u\; \}\{\; \backslash partial\; y\; \}\; =\; -\backslash frac\{\; \backslash partial\; v\; \}\{\; \backslash partial\; x\; \}$$

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

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

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

This gives the desired result

$$\backslash oint\_\backslash gamma\; f(z)\backslash ,dz\; =\; 0$$

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

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• {{springer|title=Cauchy integral theorem|id=p/c020900}}
• {{MathWorld | urlname= CauchyIntegralTheorem | title= Cauchy Integral Theorem}}
• Jeremy Orloff, 18.04 [https://ocw.mit.edu/courses/mathematics/18-04-complex-variables-with-applications-spring-2018/lecture-notes/ Complex Variables with Applications] Spring 2018 Massachusetts Institute of Technology: MIT OpenCourseWare Creative Commons.

Category:Augustin-Louis Cauchy

Category:Theorems in complex analysis