Contour integrals
Contour integral (Tag D-6DAXB)
Let \(\gamma\colon[a,b]\to\CC\) be a piecewise \(C^1\) curve and let \(f\) be a continuous complex-valued function on \(\gamma([a,b])\). The contour integral of \(f\) along \(\gamma\) is \[ \int_{\gamma}f\dz\coloneqq\int_a^b f(\gamma(t))\,\gamma'(t)\dt. \]
Write \(f=u+iv\) with \(u,v\) real-valued and \(z=x+iy\), so that \(dz=dx+i\,dy\). Then \(f\dz=(u+iv)\dx+(-v+iu)\dy\), and the contour integral is the line integral of this complex-valued \(1\)-form: \[ \int_{\gamma}f\dz=\int_\gamma (u+iv)\dx+(-v+iu)\dy. \]
Cauchy–Goursat
Cauchy's theorem for simply connected regions (Tag T-2OVCI)
Let \(\Omega\subseteq\CC\) be a simply connected region and let \(f\) be holomorphic on \(\Omega\). Then for every closed piecewise \(C^1\) curve \(\gamma\) in \(\Omega\), \[ \int_\gamma f(z)\dz=0. \]
Cauchy--Goursat theorem (Tag FT-5JQUR)
Let \(\Omega\subseteq\CC\) be a simply connected open set, let \(f\) be holomorphic on \(\Omega\), and let \(\gamma\) be a closed piecewise smooth curve in \(\Omega\). Then \[ \int_\gamma f(z)\dz = 0. \]
Closed path integrals of holomorphic functions vanish.
(Cauchy's theorem for $f$ with continuous derivative, by Stokes)
\[ \oint_{\partial D} f(z) \dz =\int_{D} d(f(z) \dz)=\int_{D}\left(\frac{\partial f}{\partial z} \dz+\frac{\partial f}{\partial \bar{z}} d \bar{z}\right) \wedge \dz=\int_{D} \frac{\partial f}{\partial \bar z} \, d \bar{z} \wedge \dz=0 ,\] since \(\dz\wedge\dz = 0\) and \(\partial f/\partial\bar z = 0\) by the Cauchy–Riemann equations.
(Goursat's theorem)
The Stokes argument assumes that \(f'\) is continuous. Goursat’s theorem proves \(\int_{\bd T} f = 0\) for every triangle \(T\) in the domain of a holomorphic \(f\) by repeated bisection, with no continuity assumption on \(f'\); continuity of \(f'\) is then a consequence of the Cauchy integral formula. The bisection proof is on Morera and converses.
Immediate consequences
The integral formula expresses \(f\) inside a curve by an integral of \(f\) over the curve.
Cauchy's estimates bound the derivatives of \(f\), and Liouville’s theorem follows.
The residue theorem computes the integral of a function with isolated singularities inside the curve.
Exercises
Primitives imply vanishing integral (Tag E-NKDKF)
Show that if \(f\) has a primitive \(F\) on \(\Omega\) then \(\int_\gamma f = 0\) for every closed curve \(\gamma \subseteq \Omega\).
Solution
Goal: Show that if \(f\) has a primitive \(F\) on \(\Omega\) (i.e. \(F' = f\) on \(\Omega\)), then \(\int_\gamma f = 0\) for every closed curve \(\gamma \subseteq \Omega\).
<1>1. If \(\gamma\) is a smooth curve from \(a\) to \(b\) parametrized by \(z(t)\), \(t \in [0,1]\), then \(\int_\gamma f = F(b) - F(a)\).
By the chain rule, \(\ddd{t}F(z(t)) = F'(z(t)) z'(t) = f(z(t)) z'(t)\), so \(\int_\gamma f = \int_0^1 f(z(t))z'(t)\,dt = \int_0^1 \ddd{t}F(z(t))\,dt = F(z(1)) - F(z(0)) = F(b) - F(a)\).
<1>2. For a closed curve \(\gamma\), the endpoints coincide, so \(\int_\gamma f = F(a) - F(a) = 0\).
<1>1 with \(b = a\): a closed curve has \(z(1) = z(0)\).
<1>3. The claim extends to piecewise smooth closed curves by summing over the smooth pieces.
Subdivide \(\gamma\) into smooth arcs; the integral is the sum of the arc integrals, and the endpoint terms telescope to \(0\) around the closed loop.
<1>4. Q.E.D.
<1>1–<1>3 prove the claim for (piecewise) smooth closed curves, which is the standard meaning of \(\int_\gamma f\) in this context.
Uniform limit theorem for holomorphic functions (Tag E-XXZVG)
Prove the uniform limit theorem for holomorphic functions: if \(f_n\to f\) locally uniformly and each \(f_n\) is holomorphic then \(f\) is holomorphic.
Solution
This is S&S Theorem 5.2. Statement: if \(f_n\to f\) uniformly locally uniformly on \(\Omega\) then \(f\) is holomorphic on \(\Omega\).
- Let \(D \subset \Omega\) with \(\bar\DD \subset \Omega\) and \(\Delta \subset D\) be a triangle.
- Apply Goursat: \(\int_\Delta f_n = 0\).
- \(f_n\to f\) uniformly on \(\Delta\) since it is closed and bounded and thus compact by Heine-Borel, so \(f\) is continuous and \[ \lim_n \int_\Delta f_n = \int_\Delta \lim_n f_n \da \int_\Delta f .\]
- Apply Morera’s theorem: \(\int_\Delta f\) vanishes on every triangle in \(\Omega\), so \(f\) is holomorphic on \(\Omega\).
Locally uniform limit theorem for holomorphic functions (Tag E-NSN6G)
Prove that if \(f_n\to f\) locally uniformly with \(f_n\) holomorphic, then \(f_n'\to f'\) locally uniformly and \(f'\) is holomorphic.
Solution
Simplifying step: for some reason, it suffices to assume \(f_n\to f\) uniformly on all of \(\Omega\)?
Take \(\Omega_R\) to be \(\Omega\) with a buffer of \(R\), so \(d(z, \bd \Omega) > R\) for every \(z \in \bar{\Omega_R}\).
It suffices to show the following bound for \(F\) any holomorphic function on \(\Omega\): \[ \sup_{z\in \Omega_R} \abs{F'(z)} \leq {1\over R} \sup_{\zeta \in \Omega} \abs{F(\zeta)} && \forall R ,\] where on the right we take the sup over all \(\Omega\).
- Then take \(F \da f_n-f\) and \(R\to 0\) to conclude, since the right-hand side is a constant not depending on \(\Omega_R\).
For any \(z\in \Omega_R\), we have \(\bar{D_R(z)} \subseteq \Omega_R\), so Cauchy’s integral formula can be applied: \[ \abs{F'(z)} &= \abs{ {1\over 2\pi i} \int_{\bd D_R(z)} {F(\xi) \over (\xi-z)^2 } \dxi } \\ &\leq {1\over 2\pi} \int_{\bd D_R(z)} { { \abs{F(\xi)} \over \abs{\xi-z}^2 }} \dxi \\ &\leq {1\over 2\pi} \int_{\bd D_R(z)} { { \sup_{\zeta\in \Omega} \abs{F(\zeta)} \over \abs{\xi-z}^2 }} \dxi \\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} \abs{F(\zeta)} \int_{\bd D_R(z)} { { 1 \over R^2 }} \dxi \\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} \abs{F(\zeta)} {1\over R^2} \int_{\bd D_R(z)} \dxi \\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} \abs{F(\zeta)} {1\over R^2} 2\pi R \\ &\leq {1\over 2\pi} \sup_{\zeta\in \Omega} \abs{F(\zeta)} {1\over R^2}\qty{ 2\pi R} \\ &= {1\over R} \sup_{\zeta \in \Omega}\abs{F(\zeta)} .\]
Now \[ \norm{f_n' - f'}_{\infty, \Omega_R} \leq {1\over R} \norm{f_n - f}_{\infty, \Omega} ,\] where if \(R\) is fixed then by uniform convergence of \(f_n\to f\), for \(n\) large enough \(\norm{f_n - f} < \eps/R\).