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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…
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 \(\Omega\), then for any simple closed contour \(C\) in \(\Omega\), 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 \(\textstyle \int_{\gamma}f(z) \, dz\) is path independent for all paths in \(U\).
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 \(U_{z_0} = \{ z : \left\vert z-z_{0}\right\vert < r\}\), for \(z_0 \in \C\), 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 \(\C\), let \(f: U \to \C\) 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 \[\int_\gamma f(z)\,dz=F(b)-F(a).\]
The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given \(U\), a simply connected open subset of \(\C\), we can weaken the assumptions to \(f\) being holomorphic on \(U\) and continuous on \(\overline{U}\) and \(\gamma\) a rectifiable simple loop in \(\textstyle \overline{U}\).
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 \[\oint_\gamma f(z)\,dz = 0 .\]
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Why is complex differentiability so much stronger than real?
The limit must be the same from every direction in the plane, not just two. That forces the Cauchy-Riemann equations, which in turn force infinitely many derivatives and a convergent Taylor series.
What is a residue?
The coefficient of 1/(z − a) in the Laurent series at a singularity a. The residue theorem says a contour integral equals 2πi times the sum of the residues inside, which evaluates many real integrals in one line.
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In ka badan Complex Analysis
The complex plane and Euler's formulaHolomorphic functions and the Cauchy-Riemann equationsContour integrals and the residue theoremPower series and analytic continuation