5.3: Proof of Cauchy's integral formula
- Page ID
- 6497
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Useful theorem
Before proving the theorem we’ll need a theorem that will be useful in its own right.
Suppose that \(A\) is a simply connected region containing the point \(z_0\). Suppose \(g\) is a function which is
- Analytic on \(A\) - {\(z_0\)}
- Continuous on \(A\). (In particular, \(g\) does not below up at \(z_0\).)
Then
\[\int_{C} g(z)\ dz = 0 \nonumber \]
for all closed curves \(C\) in \(A\).
- Proof
-
The extended version of Cauchy’s theorem in the Topic 3 notes tells us that
\[\int_C g(z)\ dz = \int_{C_r} g(z)\ dz, \nonumber \]
where \(C_r\) is a circle of radius \(r\) around \(z_0\).
Since \(g(z)\) is continuous we know that \(|g(z)|\) is bounded inside \(C_r\). Say, \(|g(z)| < M\). The corollary to the triangle inequality says that
\[|\int_{C_r} g(z)\ dz| \le M \text{ (length of } C_r) = M2 \pi r. \nonumber \]
Since \(r\) can be as small as we want, this implies that
\[\int_{C_r} g(z) \ dz = 0. \nonumber \]
Using this, we can show that \(g(z)\) is, in fact, analytic at \(z_0\). The proof will be the same as in our proof of Cauchy’s theorem that \(g(z)\) has an antiderivative.
Proof of Cauchy’s integral formula
We reiterate Cauchy’s integral formula from Equation 5.2.1: \(f(z_0) = \dfrac{1}{2\pi i} \int_C \dfrac{f(z)}{z - z_0} \ dz\).
\(Proof\). (of Cauchy’s integral formula) We use a trick that is useful enough to be worth remembering. Let
\[g(z) = \dfrac{f(z) - f(z_0)}{z - z_0}. \nonumber \]
Since \(f(z)\) is analytic on \(A\), we know that \(g(z)\) is analytic on \(A - \{z_0\}\). Since the derivative of \(f\) exists at \(z_0\), we know that
\[\lim_{z \to z_0} g(z) = f'(z_0). \nonumber \]
That is, if we define \(g(z_0) = f'(z_0)\) then \(g\) is continuous at \(z_0\). From the extension of Cauchy's theorem just above, we have
\[\int_{C} g(z)\ dz = 0, \text{ i.e. } \int_C \dfrac{f(z) - f(z_0)}{z - z_0} \ dz = 0. \nonumber \]
Thus
\[\int_{C} \dfrac{f(z)}{z - z_0}\ dz = \int_C \dfrac{f(z_0)}{z - z_0}\ dz = f(z_0) \int_C \dfrac{1}{z - z_0} \ dz = 2\pi i f(z_0). \nonumber \]
The last equality follows from our, by now, well known integral of \(1/(z - z_0)\) on a loop around \(z_0\).