<p>Integration in the Complex Plane <br>(Zill &amp; Wright Chapter 18) </p><p>1016-420-02: Complex Variables<sup style="top: -0.4338em;">∗ </sup><br>Winter 2012-2013 </p><p>Contents </p><p></p><ul style="display: flex;"><li style="flex:1">1 Contour&nbsp;Integrals </li><li style="flex:1">2</li></ul><p></p><p>2334455<br>1.1 Definition&nbsp;and Properties .&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Evaluation&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . </p><p>R</p><p></p><ul style="display: flex;"><li style="flex:1">1.2.1 Example: </li><li style="flex:1">z¯dz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . </li></ul><p></p><p>C<sub style="top: 0.0922em;">1 </sub>C<sub style="top: 0.0922em;">2 </sub></p><p>RR</p><p></p><ul style="display: flex;"><li style="flex:1">1.2.2 Example: </li><li style="flex:1">z¯dz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . </li></ul><p>1.2.3 Example:&nbsp;<sub style="top: 0.346em;">C </sub>z<sup style="top: -0.3615em;">2 </sup>dz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . <br>1.3 The&nbsp;ML Limit .&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Circulation&nbsp;and Flux&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . </p><p></p><ul style="display: flex;"><li style="flex:1">2 The&nbsp;Cauchy-Goursat Theorem </li><li style="flex:1">7</li></ul><p></p><p>789<br>2.1 Integral&nbsp;Around a Closed Loop&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Independence&nbsp;of Path for Analytic Functions .&nbsp;. . . . . . . . . . . . . . . . . 2.3 Deformation&nbsp;of Closed Contours . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 The&nbsp;Antiderivative .&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .&nbsp;10 </p><p>3 Cauchy’s&nbsp;Integral Formulas&nbsp;12 </p><p>3.1 Cauchy’s&nbsp;Integral Formula&nbsp;. . . . . . . . . . . . . . . . . . . . . . . . . . . .&nbsp;12 <br>3.1.1 Example&nbsp;#1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .&nbsp;13 3.1.2 Example&nbsp;#2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .&nbsp;13 <br>3.2 Cauchy’s&nbsp;Integral Formula for Derivatives&nbsp;. . . . . . . . . . . . . . . . . . .&nbsp;14 3.3 Consequences&nbsp;of Cauchy’s Integral Formulas&nbsp;. . . . . . . . . . . . . . . . . .&nbsp;16 <br>3.3.1 Cauchy’s&nbsp;Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . .&nbsp;16 3.3.2 Liouville’s&nbsp;Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .&nbsp;16 </p><p><sup style="top: -0.3012em;">∗</sup>Copyright 2013, John T. Whelan, and all that </p><p>1</p><p>Tuesday 18 December 2012 </p><p>1 Contour&nbsp;Integrals </p><p>1.1 Definition&nbsp;and Properties </p><p>Recall the definition of the definite integral </p><p>Z</p><p>x<sub style="top: 0.1117em;">F </sub></p><p>X</p><p></p><ul style="display: flex;"><li style="flex:1">f(x) dx = lim </li><li style="flex:1">f(x<sub style="top: 0.1494em;">k</sub>) ∆x<sub style="top: 0.1494em;">k </sub></li><li style="flex:1">(1.1) </li></ul><p></p><p>∆x<sub style="top: 0.1172em;">k</sub>→0 </p><p>x<sub style="top: 0.1116em;">I </sub></p><p>k</p><p>We’d like to define a similar concept, integrating a function f(z) from some point z<sub style="top: 0.1494em;">I </sub>to another point z<sub style="top: 0.1494em;">F </sub>. The&nbsp;problem is that, since z<sub style="top: 0.1494em;">I </sub>and z<sub style="top: 0.1494em;">F </sub>are points in the complex plane, there are different ways to get between them, and adding up the value of the function along one path will not give the same result as doing it along another path, even if they have the same endpoints. So instead we need to define a contour integral </p><p>Z<br>X</p><p></p><ul style="display: flex;"><li style="flex:1">f(z) dz = <sub style="top: 0.6642em;">|∆</sub>l<sub style="top: 0.6642em;">z</sub>im<sub style="top: 0.6642em;">|→0 </sub></li><li style="flex:1">f(z<sub style="top: 0.1494em;">k</sub>) ∆z<sub style="top: 0.1494em;">k </sub></li><li style="flex:1">(1.2) </li></ul><p></p><p>k</p><p>C</p><p>k</p><p>which in general depends on the path C by which we choose to go from the initial point z<sub style="top: 0.1494em;">I </sub>to the final point z<sub style="top: 0.1494em;">F </sub>. To define such an integral, we parametrize the curve, i.e., write it as z(t) where z(t<sub style="top: 0.1494em;">I</sub>) = z<sub style="top: 0.1494em;">I </sub>and z(t<sub style="top: 0.1494em;">F </sub>) = t<sub style="top: 0.1494em;">F </sub>. (For&nbsp;example, if C is the semi-circle in the upper half-plane, from z<sub style="top: 0.1495em;">I </sub>= 1 to z<sub style="top: 0.1495em;">F </sub>= −1, we can use the angular polar coo¨rdinate as the parameter, and then z(t) = e<sup style="top: -0.3615em;">it </sup>where t<sub style="top: 0.1495em;">I </sub>= 0 and t<sub style="top: 0.1495em;">F </sub>= π.) Given&nbsp;a parametrization of the curve, we define the contour integral as </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">t<sub style="top: 0.1117em;">F </sub></li><li style="flex:1">t<sub style="top: 0.1117em;">F </sub></li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li></ul><p></p><p>dz dt </p><ul style="display: flex;"><li style="flex:1"><sub style="top: 0.8994em;">C </sub>f(z) dz = </li><li style="flex:1">f z(t) </li><li style="flex:1">dt = </li><li style="flex:1">f z(t) z<sup style="top: -0.4113em;">0</sup>(t) dt </li></ul><p></p><p>(1.3) </p><p></p><ul style="display: flex;"><li style="flex:1">t<sub style="top: 0.1117em;">I </sub></li><li style="flex:1">t<sub style="top: 0.1116em;">I </sub></li></ul><p></p><p>You might worry that we could get a different result by choosing a different parametrization of the curve (e.g., suppose we’d chosen z(t) = e<sup style="top: -0.3615em;">it /π </sup>in the example above), but you can show </p><p>2</p><p>using the chain rule that the value of the integral doesn’t change. <br>To specify more concretely the value of the integral (1.3), write z(t) = x(t) + iy(t) and f(x+iy) = u(x, y)+iv(x, y), where x(t), y(t), u(x, y) and v(x, y) are all real functions.&nbsp;Then </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><ul style="display: flex;"><li style="flex:1">ꢂ</li><li style="flex:1">ꢃꢂ </li><li style="flex:1">ꢃ</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1"><sub style="top: 0.8994em;">C </sub>f(z) dz = </li><li style="flex:1">u(x, y) + iv(x, y) dx + idy </li></ul><p></p><p>C</p><p>Z</p><p>t<sub style="top: 0.1116em;">F </sub></p><p></p><ul style="display: flex;"><li style="flex:1">ꢂ ꢀ </li><li style="flex:1">ꢁ</li><li style="flex:1">ꢀ</li><li style="flex:1">ꢁꢃꢂ </li><li style="flex:1">ꢃ</li></ul><p></p><p>=</p><p>u x(t), y(t) +&nbsp;iv x(t), y(t) x<sup style="top: -0.4113em;">0</sup>(t) + iy<sup style="top: -0.4113em;">0</sup>(t) dt </p><p>t<sub style="top: 0.1117em;">I </sub>t<sub style="top: 0.1116em;">F </sub></p><p>Z</p><p>(1.4) </p><p></p><ul style="display: flex;"><li style="flex:1">ꢂ ꢀ </li><li style="flex:1">ꢁ</li><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢃ</li></ul><p></p><p>=</p><p>u x(t), y(t) x<sup style="top: -0.4113em;">0</sup>(t) − v x(t), y(t) y<sup style="top: -0.4113em;">0</sup>(t) dt </p><p>t<sub style="top: 0.1117em;">F </sub>t<sub style="top: 0.1117em;">I </sub></p><p>Z</p><ul style="display: flex;"><li style="flex:1">ꢂ ꢀ </li><li style="flex:1">ꢁ</li><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢃ</li></ul><p></p><p>+ i </p><p>u x(t), y(t) y<sup style="top: -0.4113em;">0</sup>(t) + v x(t), y(t) x<sup style="top: -0.4113em;">0</sup>(t) dt </p><p>t<sub style="top: 0.1116em;">I </sub></p><p>2<br>Note that Zill and Wright’s equation (2), on page 797, which says sort of the same thing, really should be written with some parentheses, to indicate that e.g., u dx and v dy are both part of the first integral. I.e., it should read </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><ul style="display: flex;"><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li></ul><p></p><p><sub style="top: 0.8994em;">C </sub>f(z) dz = u dx − v dy&nbsp;+ i v dx + u dy </p><p>(1.5) </p><p></p><ul style="display: flex;"><li style="flex:1">C</li><li style="flex:1">C</li></ul><p></p><p>I would probably take points off on a test if it were written without the parentheses! <br>Because the contour integral is defined as the limit of a sum, it has the usual linearity properties of an integral, i.e., if α and β are complex constants, </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><ul style="display: flex;"><li style="flex:1">ꢂ</li><li style="flex:1">ꢃ</li></ul><p></p><p>α f(z) + β g(z) dz = α <sub style="top: 0.8994em;">C </sub>f(z) dz + β <sub style="top: 0.8994em;">C </sub>g(z) dz </p><p>(1.6) </p><p>C</p><p>There are also contour equivalents of the properties of concatenation and reversal; just as </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">c</li><li style="flex:1">b</li><li style="flex:1">c</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">f(x) dx = </li><li style="flex:1">f(x) dx + </li><li style="flex:1">f(x) dx </li></ul><p></p><p>(1.7a) </p><p></p><ul style="display: flex;"><li style="flex:1">a</li><li style="flex:1">a</li><li style="flex:1">b</li></ul><p></p><p>and </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">a</li><li style="flex:1">b</li></ul><p></p><p>f(x) dx = − f(x) dx , </p><p>(1.7b) </p><p></p><ul style="display: flex;"><li style="flex:1">b</li><li style="flex:1">a</li></ul><p></p><p>if C<sub style="top: 0.1495em;">1 </sub>+ C<sub style="top: 0.1495em;">2 </sub>is the curve formed by chaining together C<sub style="top: 0.1495em;">1 </sub>and C<sub style="top: 0.1495em;">2 </sub>(which we can do if the final point of C<sub style="top: 0.1494em;">1 </sub>is the initial point of C<sub style="top: 0.1494em;">2</sub>), then </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">f(z) dz = </li><li style="flex:1">f(z) dz + </li><li style="flex:1">f(z) dz </li></ul><p></p><p>(1.8) </p><p></p><ul style="display: flex;"><li style="flex:1">C<sub style="top: 0.0922em;">1</sub>+C<sub style="top: 0.0922em;">2 </sub></li><li style="flex:1">C<sub style="top: 0.0922em;">1 </sub></li><li style="flex:1">C<sub style="top: 0.0922em;">2 </sub></li></ul><p></p><p>and if −C is the curve we get by following C backwards (so the initial point of −C is the final point of C and vice versa), then </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>f(z) dz = − <sub style="top: 0.8994em;">C </sub>f(z) dz </p><p>(1.9) </p><p>−C </p><p>1.2 Evaluation </p><p>R</p><p>1.2.1 Example: </p><p>z¯dz </p><p>C<sub style="top: 0.0922em;">1 </sub></p><p>R</p><p></p><ul style="display: flex;"><li style="flex:1">Consider the example of </li><li style="flex:1">z¯dz, where the function is f(z) = z¯ = x − iy, and the contour </li></ul><p></p><p>C<sub style="top: 0.0922em;">1 </sub></p><p>C<sub style="top: 0.1494em;">1 </sub>is given by x(t) = t, y(t) = t<sup style="top: -0.3615em;">2</sup>, where t<sub style="top: 0.1494em;">I </sub>= 0 and t<sub style="top: 0.1494em;">F </sub>= 1.&nbsp;Then </p><p>dz = dx + idy = [x<sup style="top: -0.4113em;">0</sup>(t) + iy<sup style="top: -0.4113em;">0</sup>(t)]dt = (1 + i 2t)dt </p><p>(1.10) (1.11) and f(z(t)) = x(t) − iy(t) = t − i t<sup style="top: -0.4113em;">2 </sup></p><p>3so </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p>ꢄ</p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>t<sub style="top: 0.1116em;">F </sub></p><p></p><ul style="display: flex;"><li style="flex:1">1</li><li style="flex:1">1</li><li style="flex:1">1</li></ul><p></p><p>z¯dz = </p><p></p><ul style="display: flex;"><li style="flex:1">f(z(t))z<sup style="top: -0.4113em;">0</sup>(t) dt = </li><li style="flex:1">(t − i t<sup style="top: -0.4113em;">2</sup>)(1 + i 2t) dt = </li><li style="flex:1">(t + 2t<sup style="top: -0.4113em;">3</sup>) dt + i </li><li style="flex:1">(−t<sup style="top: -0.4113em;">2 </sup>+ 2t<sup style="top: -0.4113em;">2</sup>) dt </li></ul><p></p><p>C<sub style="top: 0.0922em;">1 </sub></p><p>t<sub style="top: 0.1117em;">I </sub></p><p></p><ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0</li><li style="flex:1">0</li></ul><p></p><p>ꢅꢅꢅꢅ<br>ꢅꢅꢅꢅ</p><p></p><ul style="display: flex;"><li style="flex:1">1</li><li style="flex:1">1</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">t<sup style="top: -0.3616em;">2 </sup>t<sup style="top: -0.3615em;">4 </sup></li><li style="flex:1">t<sup style="top: -0.3615em;">3 </sup></li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">1</li><li style="flex:1">1</li></ul><p>2</p><ul style="display: flex;"><li style="flex:1">1</li><li style="flex:1">1</li></ul><p>3</p><ul style="display: flex;"><li style="flex:1">=</li><li style="flex:1">+</li><li style="flex:1">+ i </li><li style="flex:1">=</li><li style="flex:1">+</li><li style="flex:1">+ i = 1 + </li></ul><p></p><p>i</p><p></p><ul style="display: flex;"><li style="flex:1">2</li><li style="flex:1">2</li><li style="flex:1">3</li><li style="flex:1">2</li><li style="flex:1">3</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0</li></ul><p></p><p>(1.12) </p><p>R</p><p>1.2.2 Example: </p><p>z¯dz </p><p>C<sub style="top: 0.0922em;">2 </sub></p><p>Now let’s integrate the same integrand (f(z) = z¯ = x − iy) but along the curve C<sub style="top: 0.1494em;">2 </sub>given by x(t) = t, y(t) = t, where t<sub style="top: 0.1495em;">I </sub>= 0 and t<sub style="top: 0.1495em;">F </sub>= 1.&nbsp;Note that C<sub style="top: 0.1495em;">1 </sub>and C<sub style="top: 0.1495em;">2 </sub>both have the same endpoints z<sub style="top: 0.1494em;">I </sub>= x(t<sub style="top: 0.1494em;">I</sub>)+i y(t<sub style="top: 0.1494em;">I</sub>) = 0 and z<sub style="top: 0.1494em;">F </sub>= x(t<sub style="top: 0.1494em;">F </sub>)+i y(t<sub style="top: 0.1494em;">F </sub>) = 1+i, but the paths are different. Now </p><ul style="display: flex;"><li style="flex:1">dz = dx + idy = [x<sup style="top: -0.4113em;">0</sup>(t) + iy<sup style="top: -0.4113em;">0</sup>(t)]dt = (1 + i)dt </li><li style="flex:1">(1.13) </li></ul><p>and f(z(t)) = x(t) − iy(t) = t − i t <br>(1.14) so </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>t<sub style="top: 0.1117em;">F </sub></p><p></p><ul style="display: flex;"><li style="flex:1">1</li><li style="flex:1">1</li><li style="flex:1">1</li></ul><p></p><p>ꢅ</p><p>1<br>2</p><p>ꢅ</p><p>z¯dz = </p><p></p><ul style="display: flex;"><li style="flex:1">f(z(t))z<sup style="top: -0.4113em;">0</sup>(t) dt = </li><li style="flex:1">(t−i t)(1+i) dt = (1−i)(1+i) </li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">t dt = 2 </li><li style="flex:1">t dt = t = 1 </li></ul><p></p><p>0</p><p>C<sub style="top: 0.0922em;">1 </sub></p><p>t<sub style="top: 0.1116em;">I </sub></p><p></p><ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0</li><li style="flex:1">0</li></ul><p></p><p>(1.15) </p><p></p><ul style="display: flex;"><li style="flex:1">R</li><li style="flex:1">R</li></ul><p></p><p>so </p><p></p><ul style="display: flex;"><li style="flex:1">z¯dz = </li><li style="flex:1">z¯dz </li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">C<sub style="top: 0.0922em;">1 </sub></li><li style="flex:1">C<sub style="top: 0.0922em;">2 </sub></li></ul><p></p><p>R</p><p>1.2.3 Example:&nbsp;<sub style="top: 0.3459em;">C </sub>z<sup style="top: -0.3615em;">2 </sup>dz </p><p>As a more complicated contour, consider the integral of f(z) = z<sup style="top: -0.3615em;">2 </sup>along C, which is a semicircle of radius 2, from z<sub style="top: 0.1494em;">I </sub>= 2 to z<sub style="top: 0.1494em;">F </sub>= −2, counter-clockwise in the upper half plane. <br>Now we have to start by making a parametrization of the curve.&nbsp;Different choices are possible, but a convenient one is </p><p></p><ul style="display: flex;"><li style="flex:1">z = 2 e<sup style="top: -0.4114em;">it </sup></li><li style="flex:1">with </li><li style="flex:1">t<sub style="top: 0.1494em;">I </sub>= 0&nbsp;and t<sub style="top: 0.1494em;">F </sub>= π </li><li style="flex:1">(1.16) </li></ul><p>Rather than splitting things up into u(x, y), v(x, y), etc, it’s more straightforward to work with z(t) and write </p><p>dz = z<sup style="top: -0.4113em;">0</sup>(t) dt = 2i e<sup style="top: -0.4113em;">it </sup>dt </p><p>(1.17) (1.18) </p><p>ꢁ</p><p>and f(z(t)) = (2e<sup style="top: -0.4114em;">it</sup>)<sup style="top: -0.4114em;">2 </sup>= 4e<sup style="top: -0.4114em;">i2t </sup></p><p>so </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>t<sub style="top: 0.1116em;">F </sub></p><p></p><ul style="display: flex;"><li style="flex:1">π</li><li style="flex:1">π</li></ul><p></p><p>ꢅꢅ<br>ꢀ</p><p>8</p><p>8i e<sup style="top: -0.4113em;">i3t </sup>dt = e<sup style="top: -0.4113em;">i3t </sup></p><p>3<br>83</p><p>π</p><p><sub style="top: 0.8994em;">C </sub>z¯dz = </p><p>f(z(t))z<sup style="top: -0.4113em;">0</sup>(t) dt = </p><p>4e<sup style="top: -0.4113em;">i2t</sup>2i e<sup style="top: -0.4113em;">it </sup>dt = </p><p>=</p><p>e<sup style="top: -0.4113em;">i3π </sup>− e<sup style="top: -0.4113em;">0 </sup></p><p>0</p><p>t<sub style="top: 0.1117em;">I </sub></p><p></p><ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0</li></ul><p></p><p>83<br>16 <br>=<br>(−1 − 1) = − <br>3<br>(1.19) </p><p>4</p><p>1.3 The&nbsp;ML Limit </p><p>It is occasionally useful to place an upper limit on a contour integral, rather than evaluating it. A&nbsp;limit that can be placed on any contour integral of a continuous function along a smooth curve is the so-called ML limit: </p><p></p><ul style="display: flex;"><li style="flex:1">ꢅ</li><li style="flex:1">ꢅ</li></ul><p>Z</p><p>ꢅꢅꢅ<br>ꢅꢅ</p><p>f(z) dz ≤ ML </p><p>(1.20) </p><p>ꢅ</p><p>C</p><p>where M is the maximum value of |f(z)|: </p><p>|f(z)| ≤ M </p><p></p><ul style="display: flex;"><li style="flex:1">for all z on C </li><li style="flex:1">(1.21) </li></ul><p>and L is the length of C. </p><p>1.4 Circulation&nbsp;and Flux </p><p>The contour integral also appears in vector calculus, where can evaluate the integral of a scalar field Ψ(~r) along some path ~r(t) as </p><p>ꢅꢅꢅꢅ<br>ꢅꢅꢅꢅ</p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>t<sub style="top: 0.1117em;">F </sub></p><p>d~r dt </p><p></p><ul style="display: flex;"><li style="flex:1"><sub style="top: 0.8994em;">C </sub>Ψ(~r) ds = </li><li style="flex:1">Ψ (~r(t)) </li></ul><p></p><p>dt , </p><p>(1.22) </p><p>t<sub style="top: 0.1116em;">I </sub></p><p>where </p><p>r<br>ꢅ</p><p>ꢅꢅꢅ<br>ꢅꢅꢅꢅ</p><p>d~r dt d~r d~r </p><p>·</p><p>=</p><p>.</p><p>(1.23) (1.24) </p><p>dt dt </p><p>In two dimensions, i.e., </p><p>p</p><p>ds = (dx)<sup style="top: -0.2878em;">2 </sup>+ (dy)<sup style="top: -0.2878em;">2 </sup></p><p>s<br>ꢅꢅꢅꢅ<br>ꢅꢅꢅꢅ</p><ul style="display: flex;"><li style="flex:1">ꢆ</li><li style="flex:1">ꢇ</li><li style="flex:1">ꢆ</li><li style="flex:1">ꢇ</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">2</li><li style="flex:1">2</li></ul><p></p><p>d~r dt dx dt dy dt </p><p></p><ul style="display: flex;"><li style="flex:1">=</li><li style="flex:1">+</li><li style="flex:1">(1.25) </li></ul><p>(1.26) <br>Given a vector field </p><p></p><ul style="display: flex;"><li style="flex:1">~</li><li style="flex:1">~</li></ul><p>F(~r) = F(x, y) = F<sub style="top: 0.1494em;">x</sub>(x, y) xˆ + F<sub style="top: 0.1494em;">y</sub>(x, y) yˆ , </p><p>and a smooth closed curve C, we’re often interested in two contour integrals, known as the </p><p></p><ul style="display: flex;"><li style="flex:1">~</li><li style="flex:1">~</li></ul><p></p><p>flux of F through C and the circulation of F around C. At&nbsp;each point on the closed curve <br>ˆ<br>C we define a unit vector t pointing along the curve in the counter-clockwise direction and </p><p>another unit vector nˆ pointing perpendicular to the curve in the outward direction.&nbsp;Then the integrals of interest are </p><p>Z</p><p></p><ul style="display: flex;"><li style="flex:1">~</li><li style="flex:1">~</li></ul><p></p><p>ˆ</p><ul style="display: flex;"><li style="flex:1">Circulation of F around C = F · t ds </li><li style="flex:1">(1.27a) </li></ul><p>(1.27b) </p><p>C</p><p>Z</p><p></p><ul style="display: flex;"><li style="flex:1">~</li><li style="flex:1">~</li></ul><p></p><p>Flux of F through C = F · nˆ ds </p><p>C</p><p>5where the arrow on the integral sign emphasizes that C is a closed curve traversed counterclockwise. A little bit of geometry shows </p><p>ˆ</p><p>t ds = xˆ dx + yˆdy </p><p>(1.28a) (1.28b) </p><p>nˆ ds = xˆ dy − yˆdx </p><p>so </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>~</p><p>ˆ<br>Circulation =&nbsp;F · t ds = [F<sub style="top: 0.1494em;">x </sub>dx + F<sub style="top: 0.1494em;">y </sub>dy] <br>(1.29a) (1.29b) </p><p></p><ul style="display: flex;"><li style="flex:1">C</li><li style="flex:1">C</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>~</p><p>Flux =&nbsp;F · nˆ ds = [F<sub style="top: 0.1494em;">x </sub>dy − F<sub style="top: 0.1494em;">y </sub>dx] </p><p></p><ul style="display: flex;"><li style="flex:1">C</li><li style="flex:1">C</li></ul><p></p><p>Now, consider our friend the P´olya vector field associated with a function f(z) = f(x+iy) = u(x, y) + iv(x, y), defined </p><p>~H = u(x, y) xˆ − v(x, y)yˆ </p><p>(1.30) with the minus sign on the y component. Its&nbsp;circulation around and flux through some closed curve C are </p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>~</p><p>Circulation of H around C = [H<sub style="top: 0.1494em;">x </sub>dx + H<sub style="top: 0.1494em;">y </sub>dy] =&nbsp;[u dx − v dy] <br>(1.31a) (1.31b) </p><p></p><ul style="display: flex;"><li style="flex:1">C</li><li style="flex:1">C</li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1">Z</li><li style="flex:1">Z</li></ul><p></p><p>~</p><p>Flux of H through C = [H<sub style="top: 0.1495em;">x </sub>dy − H<sub style="top: 0.1495em;">y </sub>dx] =&nbsp;<sub style="top: 0.8994em;">C </sub>[u dy + v dx] </p><p>C</p><p>But if we compare this to (1.4) or (1.5), we see that these are just the real and imaginary </p><p>R</p><p>parts of the contour integral&nbsp;<sub style="top: 0.346em;">C </sub>f(z) dz! I.e., </p><p></p><ul style="display: flex;"><li style="flex:1">ꢆ</li><li style="flex:1">ꢇ</li></ul><p>ꢇ<br>Z</p><p>~</p><p></p><ul style="display: flex;"><li style="flex:1">Circulation of H = uxˆ − vyˆ around C = Re&nbsp;<sub style="top: 0.8994em;">C </sub>f(z) dz </li><li style="flex:1">(1.32a) </li></ul><p>(1.32b) </p>