4.5: Examples
- Page ID
- 6484
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Why can't we compute \(\int_{\gamma} \overline{z}\ dz\) using the fundamental theorem.
Solution
Because \(\overline{z}\) doesn’t have an antiderivative. We can also see this by noting that if \(\overline{z}\) had an antiderivative, then its integral around the unit circle would have to be 0. But, we saw in Example 4.2.4 that this is not the case.
Compute \(\int_{\gamma} \dfrac{1}{z}\ dz\) over each of the following contours
- The line from 1 to \(1 + i\).
- The circle of radius 1 around \(z = 3\).
- The unit circle.
Solution
For parts (i) and (ii) there is no problem using the antiderivative \(\log (z)\) because these curves are contained in a simply connected region that doesn’t contain the origin.
(i)
\[\int_{\gamma} \dfrac{1}{z}\ dz = \log (1 + i) - \log (1) = \log (\sqrt{2}) + i \dfrac{\pi}{4}. \nonumber \]
(ii) Since the beginning and end points are the same, we get
\[\int_{\gamma} \dfrac{1}{z} \ dz = 0 \nonumber \]
(iii) We parametrize the unit circle by \(\gamma (\theta) = e^{i \theta}\) with \(0 \le \theta \le 2\pi\). We compute \(\gamma '(\theta) = ie^{i \theta}\). So the integral becomes
\[\int_{\gamma} \dfrac{1}{z} \ dz = \int_{0}^{2\pi} \dfrac{1}{e^{i \theta}} ie^{i \theta} \ dt = \int_{0}^{2\pi} i \ dt = 2\pi i. \nonumber \]
Notice that we could use \(\log (z)\) if we were careful to let the argument increase by \(2 \pi\) as it went around the origin once.
Compute \(\int_{\gamma} \dfrac{1}{z^2} \ dz\), where \(\gamma\) is the unit circle in two ways.
- Using the fundamental theorem.
- Directly from the definition.
Solution
(i) Let \(f(z) = -1/z\). Since \(f'(z) = 1/z^2\), the fundamental theorem says
\[\int_{\gamma} \dfrac{1}{z^2} \ dz = \int_{\gamma} f'(z) \ dz = f(\text{endpint}) - f(\text{start point}) = 0. \nonumber \]
It equals 0 because the start and endpoints are the same.
(ii) As usual, we parametrize the unit circle as \(\gamma (\theta = e^{i \theta}\) with \(0 \le \theta \le 2\pi\). So, \(\gamma '(\theta) = ie^{i \theta}\) and the integral becomes
\[\int_{\gamma} \dfrac{1}{z^2} \ dz = \int_{0}^{2 \pi} \dfrac{1}{e^{2i \theta}} ie^{i \theta}\ d \theta = \int_{0}^{2\pi} i e^{-i \theta}\ d \theta = -e^{-i \theta} \vert_{0}^{2\pi} = 0. \nonumber \]