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# SSM-Ch14 - Chap 14 Complex Integration This is the first...

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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 ue0a2 t ue0a3 u003d x ue0a2 t ue0a3 u002b iy ue0a2 t ue0a3 u003d ue0a2 1 u002b 3 i ue0a3 t u003d t u002b 3 it is linear in t , this is a straight line in the complex z -plane. Its slope is positive, y ue0a2 t ue0a3 / x ue0a2 t ue0a3 u003d 3. The line passes through the origin. t u003d 1 and t u003d 4 correspond to the points 1 u002b 3 i and ue0a2 1 u002b 3 i ue0a3 u2219 4 u003d 4 u002b 12 i . These are the endpoints of the straight line segment represented by the given data. Sketch it. 7 . Ellipse . From z ue0a2 t ue0a3 u003d x ue0a2 t ue0a3 u002b iy ue0a2 t ue0a3 u003d 6 cos 2 t u002b 5 i sin 2 t you obtain x ue0a2 t ue0a3 u003d 6 cos 2 t and y ue0a2 t ue0a3 u003d 5 sin 2 t and thus x 6 2 u002b y 5 2 u003d x 2 6 2 u002b y 2 5 2 u003d cos 2 2 t u002b sin 2 2 t u003d 1. This is an ellipse with semiaxes 6 and 5. t varies from 0 to u03c0 as given, so that 2 t varies from 0 to 2 u03c0 ; hence the representation includes the whole ellipse. 13 . Parametric representation . You have y u003d 1/ x and may thus set x u003d t , y u003d 1/ t . Now z 0 u003d 1 u002b i corresponds to t u003d 1, and z 1 u003d 4 u002b 1 4 i corresponds to t u003d 4. Hence you obtain z ue0a2 t ue0a3 u003d x ue0a2 t ue0a3 u002b iy ue0a2 t ue0a3 u003d t u002b i / t ue0a2 1 u2264 t u2264 4 ue0a3 . 19 . Integration . Re z u003d x is not analytic. Hence integrate by the use of the path. Find a representation of the path C , z ue0a2 t ue0a3 u003d ue0a2 1 u002b i ue0a3 t u003d t u002b it , 0 u2264 t u2264 1. Since Re z ue0a2 t ue0a3 u003d t and dz / dt u003d 1 u002b i , you obtain u222b C Re z dz u003d u222b 0 1 t ue0a2 1 u002b i ue0a3 dt u003d 1 2 t 2 ue0a2 1 u002b i ue0a3 0 1 u003d 1 2 ue0a2 1 u002b i ue0a3 . 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 u003d 1 2 u002b 1 2 cos 2 z (see (10) on p. A61). Integrate this to obtain (with a factor 1/2 from the chain rule)

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138 Complex Analysis Part D 1 2 z u002b 1 4 sin 2 z u2212 u03c0 i u03c0 i u003d 1 2 ue0a2 u03c0 i u2212 ue0a2 u2212 u03c0 i ue0a3ue0a3 u002b 1 4 sin 2 u03c0 i u2212 1 4 sin ue0a2 u2212 2 u03c0 i ue0a3 u003d u03c0 i u002b 1 2 sin 2 u03c0 i ue0a4 now use ue0a2 15 ue0a3 , Sec.13.6 ue0a5 u003d u03c0 i u002b 1 2 i sinh 2 u03c0 . 27 . First method . Integrate sec 2 z u003d 1/cos 2 z to obtain tan z , as in calculus. Insert the limits of integration , obtaining tan 1 4 u03c0 i u2212 tan 1 4 u03c0 . The second term equals u2212 1 (calculus!). For the first term use (15) in Sec. 13.6 and the definition of the hyperbolic tangent: sin 1 4 u03c0 i cos 1 4 u03c0 i u003d i sinh 1 4 u03c0 cosh 1 4 u03c0 u003d i tanh 1 4 u03c0 , where the last equality follows directly from the definition of the hyperbolic tangent. A numeric value (6S) is 0.655794 i . Remember that the real hyperbolic tangent varies between
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SSM-Ch14 - Chap 14 Complex Integration This is the first...

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