Chapter 18

# Chapter 18 - 18 Integration in the Complex Plane Exercises...

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Unformatted text preview: 18 Integration in the Complex Plane Exercises 18.1 1. Z C ( z + 3) dz = (2 + 4 i ) · Z 3 1 (2 t + 3) dt + i Z 3 1 (4 t − 1) dt ¸ = (2 + 4 i )[14 + 14 i ] = − 28 + 84 i 2. Z C (2¯ z − z ) dz = Z 2 [ − t − 3( t 2 + 2) i ]( − 1 + 2 ti ) dt = Z 2 (6 t 3 + 13 t ) dt + i Z 2 ( t 2 + 2) dt = 50 + 20 3 i 3. Z C z 2 dz = (3 + 2 i ) 3 Z 2 − 2 t 2 dt = 16 3 (3 + 2 i ) 3 = − 48 + 736 3 i 4. Z C (3 z 2 − 2 z ) dz = Z 1 ( − 15 t 4 + 4 t 3 + 3 t 2 − 2 t ) dt + i Z 1 ( − 6 t 5 + 12 t 3 − 6 t 2 ) dt = − 2 + 0 i = − 2 5. Using z = e it , − π/ 2 ≤ t ≤ π/ 2, and dz = ie it dt , Z C 1 + z z dz = − Z π/ 2 − π/ 2 (1 + e it ) dt = (2 + π ) i . 6. Z C | z | 2 dz = Z 2 1 µ 2 t 5 + 2 t ¶ dt − i Z 2 1 µ t 2 + 1 t 4 ¶ dt = 21 + ln4 − 21 8 i 7. Using z = e it = cos t + i sin t , dz = ( − sin t + i cos t ) dt and x = cos t , I ˇ C Re( z ) dz = Z 2 π cos t ( − sin t + i cos t ) dt = − Z 2 π sin t cos t dt + i Z 2 π cos 2 t dt = − 1 2 Z 2 π sin2 t dt + 1 2 i Z 2 π (1 + cos2 t ) dt = πi. 8. Using z + i = e it , 0 ≤ t ≤ 2 π , and dz = ie it dt , I ˇ C · 1 ( z + i ) 3 − 5 z + i + 8 ¸ dz = i Z 2 π [ e − 2 it − 5 + 8 e it ] dt = − 10 πi. 9. Using y = − x + 1, 0 ≤ x ≤ 1, z = x + ( − x + 1) i , dz = (1 − i ) dx , Z C ( x 2 + iy 3 ) dz = (1 − i ) Z 1 [ x 2 + (1 − x ) 3 i ] dx = − 7 12 + 1 12 i. 10. Using z = e it , π ≤ t ≤ 2 π , dz = ie it dt , x = cos t = ( e it + e − it ) / 2, y = sin t = ( e it − e − it ) / 2 i , Z C ( x 3 − iy 3 ) dz = 1 8 i Z 2 π π ( e 3 it + 3 e it + 3 e − it + e − 3 it ) e it dt + 1 8 i Z 2 π π ( e 3 it − 3 e it + 3 e − it − e − 3 it ) e it dt = 1 8 i Z 2 π π (2 e 4 it + 6) dt = 3 π 4 i. 11. Z C e z dz = Z C 1 e z dz + Z C 2 e z dz where C 1 and C 2 are the line segments y = 0, 0 ≤ x ≤ 2 and y = − πx + 2 π , 1 ≤ x ≤ 2, respectively. Now Z C 1 e z dz = Z 2 e x dx = e 2 − 1 Z C 2 e z dz = (1 − πi ) Z 1 2 e x +( − πx +2 π ) i dx = (1 − πi ) Z 1 2 e (1 − πi ) x dx = e 1 − πi − e 2(1 − πi ) = − e − e 2 . 779 Exercises 18.1 In the second integral we have used the fact that e z has period 2 πi . Thus Z C e z dz = ( e 2 − 1) + ( − e − e 2 ) = − 1 − e. 12. Z C sin z dz = Z C 1 sin z dz + Z C 2 sin z dz where C 1 and C 2 are the line segments y = 0, 0 ≤ x ≤ 1, and x = 1, ≤ y ≤ 1, respectively. Now Z C 1 sin z dz = Z 1 sin x dx = 1 − cos1 Z C 2 sin z dz = i Z 1 sin(1 + iy ) dy = cos1 − cos(1 + i ) . Thus Z C sin z dz = (1 − cos1)+(cos1 − cos(1+ i )) = 1 − cos(1+ i ) = (1 − cos1cosh1)+ i sin1sinh1 = 0 . 1663+0 . 9889 i. 13. We have Z C Im( z − i ) dz = Z C 1 ( y − 1) dz + Z C 2 ( y − 1) dz On C 1 , z = e it , 0 ≤ t ≤ π/ 2, dz = ie it dt , y = sin t = ( e it − e − it ) / 2 i , Z C 1 = ( y − 1) dz = 1 2 Z π/ 2 [ e it − e − it − 2 i ] e it dt = 1 2 Z π/ 2 [ e 2 it − 1 + 2 ie it ] dt = 1 − π 4 − 1 2 i. On C 2 , y = x + 1, − 1 ≤ x ≤ 0, z = x + ( x + 1) i , dz = (1 + i ) dx , Z C 2 ( y − 1) dz = (1 + i ) Z − 1 x dx = 1 2 + 1 2 i. Thus Z C Im( z − i ) dz = µ 1 − π 4 − 1 2 i ¶ + µ 1 2 + 1 2 i ¶ = 3 2 − π 4 . 14. Using x = 6cos t , y = 2sin t , π/ 2 ≤ t ≤ 3 π/ 2, z = 6cos t + 2 i sin t , dz = ( − 6sin t + 2 i cos t ) dt , Z C dz = − 6 Z 3 π/ 2 π/ 2 sin t dt + 2 i Z 3 π/ 2 π/ 2 cos t dt = 2 i ( − 2) = − 4 i....
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## This note was uploaded on 02/13/2008 for the course ENG 342 taught by Professor Delahanty during the Fall '07 term at TCNJ.

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Chapter 18 - 18 Integration in the Complex Plane Exercises...

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