380_sample_exam_2_solns[1]

380_sample_exam_2_solns[1] - Solutions to the sample exam 2...

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Solutions to the sample exam 2 problems 1. (a) Here, the path 1 is parametrized by z = (2 + i ) t , with 0 t 1 : So z = (2 ± i ) t and dz = (2 + i ) dt: Thus Z 1 zdz = Z 1 0 (2 ± i ) t (2 + i ) dt = 5 2 : (b) Here, the curve 2 is the reverse of the curve parametrized by z = 1 + 2 e it ; 0 t ± for which z = 1 + 2 e it and dz = 2 ie it dt: Thus Z 2 zdz = ± Z ± 0 1 + 2 e it ± (2 ie it ) dt = 4 ± 4 (c) Let ² join 0 to i and ³ join i to 2 + i: Then ² is parametrized by z = iy; 0 y 1 : The curve ³ is given by z = x + i , 0 x 2 : Thus Z ² + ³ zdz = Z ² zdz + Z ³ zdz = Z 1 0 ( ± iy ) idy + Z 2 0 ( x ± i ) dx = 5 2 ± 2 i: 2. In this problem, the contour integrals are path independent since e 2 z has a global anti-derivative, F ( z ) = 1 2 e 2 z : Thus (a) R 1 e 2 z dz = F (2 + i ) ± F (0) = 1 2 e 4+2 i ± 1 2 e 0 = 1 2 e 4+2 i ± 1 ± (b) R 2 e 2 z dz = F (3) ± f ( ± 1) = 1 2 ( e 6 ± e 2 ) (c) same as 2(a). 3. The sets in (b) and (c) are simply connected. The set in (a) fails to be
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380_sample_exam_2_solns[1] - Solutions to the sample exam 2...

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