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Unformatted text preview: Chap . 14 Complex Integration This is the first chapter on complex integration. It explains the evaluation of complex line integrals, the analogs of real line integrals. Another chapter on complex integration is Chap. 16, which is based on infinite series, whose discussion begins in Chap. 15. Sec . 14 . 1 Line Integral in the Complex Plane This section contains the definition, basic properties, the existence, and two integration methods. Problem Set 14 . 1 . Page 645 1 . Parametric representation . Since z u t U u x u t U U iy u t U u u 1 U 3 i U t u t U 3 it is linear in t , this is a straight line in the complex zplane. Its slope is positive, y u t U / x u t U u 3. The line passes through the origin. t u 1 and t u 4 correspond to the points 1 U 3 i and u 1 U 3 i U u 4 u 4 U 12 i . These are the endpoints of the straight line segment represented by the given data. Sketch it. 7 . Ellipse . From z u t U u x u t U U iy u t U u 6 cos 2 t U 5 i sin 2 t you obtain x u t U u 6 cos 2 t and y u t U u 5 sin 2 t and thus x 6 2 U y 5 2 u x 2 6 2 U y 2 5 2 u cos 2 2 t U sin 2 2 t u 1. This is an ellipse with semiaxes 6 and 5. t varies from 0 to u as given, so that 2 t varies from 0 to 2 u ; hence the representation includes the whole ellipse. 13 . Parametric representation . You have y u 1/ x and may thus set x u t , y u 1/ t . Now z u 1 U i corresponds to t u 1, and z 1 u 4 U 1 4 i corresponds to t u 4. Hence you obtain z u t U u x u t U U iy u t U u t U i / t u 1 U t U 4 U . 19 . Integration . Re z u x is not analytic. Hence integrate by the use of the path. Find a representation of the path C , z u t U u u 1 U i U t u t U it , U t U 1. Since Re z u t U u t and dz / dt u 1 U i , you obtain ¡ C Re z dz u ¡ 1 t u 1 U i U dt u 1 2 t 2 u 1 U i U 1 u 1 2 u 1 U i U . 23 . Integration by the first method . cos 2 z is analytic. Hence use indefinite integration and substitution of limits. As in calculus, use cos 2 z u 1 2 U 1 2 cos 2 z (see (10) on p. A61). Integrate this to obtain (with a factor 1/2 from the chain rule) 138 Complex Analysis Part D 1 2 z u 1 4 sin 2 z u u i u i U 1 2 u u i u u u u i UU u 1 4 sin 2 u i u 1 4 sin u u 2 u i U U u i u 1 2 sin 2 u i ¡ now use u 15 U , Sec.13.6 ¢ U u i u 1 2 i sinh 2 u . 27 . First method . Integrate sec 2 z U 1/cos 2 z to obtain tan z , as in calculus. Insert the limits of integration , obtaining tan 1 4 u i u tan 1 4 u ....
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.
 Spring '12
 M.lee

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