# HW6 - MATH 185 SOLUTIONS OF HW VI(1(40 1(a By parametrizing...

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MATH 185 SOLUTIONS OF HW VI (1) (40. 1) (a) By parametrizing z = 2 e (then dz = 2 ie ), Z C z + 2 z dz = Z π 0 ± 2 e + 2 2 e ² 2 ie = 2 i Z π 0 ( e +1) = 2 i [ - ie + θ ] π 0 = - 4+2 πi (b) Doing similarly, we get Z C z + 2 z dz = Z 2 π π ± 2 e + 2 2 e ² 2 ie = 2 i Z 2 π π ( e +1) = 2 i [ - ie + θ ] 2 π π = 4+2 πi (c) Since C = C a + C b where C a is the semicircle in (a), C b is the semicircle in (b), we have Z C f ( z ) dz = Z C a f ( z ) dz + Z C b f ( z ) dz = ( - 4 + 2 πi ) + (4 + 2 πi ) = 4 πi (2) (40. 3) Let I 1 ( I 2 , I 3 , I 4 ) be the line segment from 0 to 1(from 1 to 1+ i , from 1+ i to i , from i to 0, respectively). On I 1 , we can parametrize z = x, f ( z ) = πe πx . Hence Z I 1 f ( z ) dz = Z 1 0 πe πx dx = [ e πx ] 1 0 = e π - 1 Similarly, on I 2 , we can parametrize z = 1 + iy, f ( z ) = πe (1 - iy ) π . Hence Z I 2 f ( z ) dz = Z 1 0 πe (1 - iy ) π idy = [ - e (1 - iy ) π ] 1 0 = - e (1 - i ) π + e π = 2

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HW6 - MATH 185 SOLUTIONS OF HW VI(1(40 1(a By parametrizing...

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