Solutions_MT2_Test_fall2018_185.pdf - Math 185 \u2014 Solutions to Test Midterm 2 Question 1(2 2 3 3 points Let C be a positively oriented simple closed

Solutions_MT2_Test_fall2018_185.pdf - Math 185 —...

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Math 185 — Solutions to Test Midterm 2 October 24, 2018 Question 1 (2+2+3+3 points) . Let C be a positively oriented simple closed contour sur- rounding the origin. Compute the following four integrals and give reason for each step in your computation. (1) Z C dz z Solution: Since the integrand is analytic in C \ { 0 } , by the deformation of paths theorem we may replace C by the positively oriented unit circle | z | = 1, which is parametrized by z ( t ) = e it (0 t 2 π ). Then z 0 ( t ) = ie it and therefore Z C dz z = Z | z | =1 dz z by def = Z 2 π 0 1 e it ie it dt = Z 2 π 0 i dt = 2 πi. Alternatively, applying Cauchy’s integral formula to the entire function f ( z ) = 1, we obtain Z C dz z = 2 πif (0) = 2 πi. (2) Z C dz z 2 Solution: Since the integrand is again analytic in C \{ 0 } we can argue as in (1) to deform the path and obtain Z C dz z 2 = Z | z | =1 dz z 2 by def = Z 2 π 0 1 e 2 it ie it dt = Z 2 π 0 d dt ( - e - it ) dt = 0 . Alternatively, since F ( z ) = - 1 z is analytic in C \ { 0 } with F 0 ( z ) = 1 z 2 we obtain Z C dz z 2 = 0 . 1
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Yet another alternative is to apply the extended Cauchy integral formula (for the first derivative) to the entire function f ( z ) = 1, so that Z C dz z 2 = 2 πif 0 (0) = 0 . (3) Z C 1 z exp(sin( z )) dz Solution: By the chain rule, the function f ( z ) = exp(sin( z )) is entire. Therefore, we may apply Cauchy’s integral formula: Z C 1 z exp(sin( z )) dz = Z C f ( z ) z dz = 2 πif (0) = 2 πi. (4) Z C z 3 exp(5 πz ) dz Solution: By the product rule, the integrand is entire, so by the Cauchy-Goursat theorem any integral along a closed contour vanishes, in particular Z C z 3 exp(5 πz ) dz = 0 . Question 2 (4+6 points) . 1. Let C be a contour and suppose that a sequence ( f n ) n of continuous functions converges uniformly on C to some function f . Show that the integrals R C f n ( z ) dz converge to R C f ( z ) dz . Solution: Let ε > 0 and let L denote the length of the contour C . We want to show that there exists an n 0 such that for all n n 0 Z C f ( z ) dz - Z C f n ( z ) dz < ε. In order to prove this statement, we observe that by assumption we can find n 0 N such that for all n n 0 and all z C it holds | f ( z ) - f n ( z ) | < ε L . Hence by the theorem on the modulus of an integral, Z C f ( z ) dz - Z
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