5.1: Cauchy's Integral for Functions
- Page ID
- 6495
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Suppose \(C\) is a simple closed curve and the function \(f(z)\) is analytic on a region containing \(C\) and its interior (Figure \(\PageIndex{1}\)). We assume \(C\) is oriented counterclockwise.
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Then for any \(z_0\) inside \(C\):
\[f(z_0) = \dfrac{1}{2\pi i} \int_C \dfrac{f(z)}{z - z_0} \ dz \nonumber \]
This is remarkable: it says that knowing the values of \(f\) on the boundary curve \(C\) means we know everything about \(f\) inside \(C\)!! This is probably unlike anything you’ve encountered with functions of real variables.
Aside 1. With a slight change of notation (\(z\) becomes \(w\) and \(z_0\) becomes \(z\)) we often write the formula as
\[f(z) = \dfrac{1}{2 \pi i} \int_C \dfrac{f(w)}{w - z} \ dw \nonumber \]
Aside 2. We’re not being entirely fair to functions of real variables. We will see that for \(f = u + iv\) the real and imaginary parts \(u\) and \(v\) have many similar remarkable properties. \(u\) and \(v\) are called conjugate harmonic functions.
Compute \(\int_c \dfrac{e^{z^2}}{z - 2} \ dz\), where \(C\) is the curve shown in Figure \(\PageIndex{2}\).
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Solution
Let \(f(z) = e^{z^2}\). \(f(z)\) is entire. Since \(C\) is a simple closed curve (counterclockwise) and \(z = 2\) is inside \(C\), Cauchy’s integral formula says that the integral is \(2 \pi i f(2) = 2\pi i e^4\).
Do the same integral as the previous example with \(C\) the curve shown in Figure \(\PageIndex{3}\).
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Solution
Since \(f(z) = e^{z^2} / (z - 2)\) is analytic on and inside \(C\), Cauchy’s theorem says that the integral is 0.
Do the same integral as the previous examples with \(C\) the curve shown.
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Solution
This one is trickier. Let \(f(z) = e^{z^2}\). The curve \(C\) goes around 2 twice in the \(clockwise\) direction, so we break \(C\) into \(C_1 + C_2\) as shown in the next figure.
.svg?revision=1&size=bestfit&width=423&height=279)
These are both simple closed curves, so we can apply the Cauchy integral formula to each separately. (The negative signs are because they go clockwise around \(z = 2\).)
\[\int_C \dfrac{f(z)}{z - 2} \ dz = \int_{C_1} \dfrac{f(z)}{z - 2} \ dz + \int_{C_2} \dfrac{f(z)}{z - 2} \ dz = -2\pi i f(2) - 2\pi i f(2) = -4\pi i f(2). \nonumber \]