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 ) Β· Z 3 1 (2 t + 3) dt + i Z 3 1 (4 t βˆ’ 1) dt ΒΈ = (2 + 4 i )[14 + 14 i ] = βˆ’ 28 + 84 i 2. Z C (2Β― z βˆ’ z ) dz = Z 2 [ βˆ’ t βˆ’ 3( t 2 + 2) i ]( βˆ’ 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 βˆ’ 2 t 2 dt = 16 3 (3 + 2 i ) 3 = βˆ’ 48 + 736 3 i 4. Z C (3 z 2 βˆ’ 2 z ) dz = Z 1 ( βˆ’ 15 t 4 + 4 t 3 + 3 t 2 βˆ’ 2 t ) dt + i Z 1 ( βˆ’ 6 t 5 + 12 t 3 βˆ’ 6 t 2 ) dt = βˆ’ 2 + 0 i = βˆ’ 2 5. Using z = e it , βˆ’ Ο€/ 2 ≀ t ≀ Ο€/ 2, and dz = ie it dt , Z C 1 + z z dz = βˆ’ Z Ο€/ 2 βˆ’ Ο€/ 2 (1 + e it ) dt = (2 + Ο€ ) i . 6. Z C | z | 2 dz = Z 2 1 Β΅ 2 t 5 + 2 t ΒΆ dt βˆ’ i Z 2 1 Β΅ t 2 + 1 t 4 ΒΆ dt = 21 + ln4 βˆ’ 21 8 i 7. Using z = e it = cos t + i sin t , dz = ( βˆ’ sin t + i cos t ) dt and x = cos t , I Λ‡ C Re( z ) dz = Z 2 Ο€ cos t ( βˆ’ sin t + i cos t ) dt = βˆ’ Z 2 Ο€ sin t cos t dt + i Z 2 Ο€ cos 2 t dt = βˆ’ 1 2 Z 2 Ο€ sin2 t dt + 1 2 i Z 2 Ο€ (1 + cos2 t ) dt = Ο€i. 8. Using z + i = e it , 0 ≀ t ≀ 2 Ο€ , and dz = ie it dt , I Λ‡ C Β· 1 ( z + i ) 3 βˆ’ 5 z + i + 8 ΒΈ dz = i Z 2 Ο€ [ e βˆ’ 2 it βˆ’ 5 + 8 e it ] dt = βˆ’ 10 Ο€i. 9. Using y = βˆ’ x + 1, 0 ≀ x ≀ 1, z = x + ( βˆ’ x + 1) i , dz = (1 βˆ’ i ) dx , Z C ( x 2 + iy 3 ) dz = (1 βˆ’ i ) Z 1 [ x 2 + (1 βˆ’ x ) 3 i ] dx = βˆ’ 7 12 + 1 12 i. 10. Using z = e it , Ο€ ≀ t ≀ 2 Ο€ , dz = ie it dt , x = cos t = ( e it + e βˆ’ it ) / 2, y = sin t = ( e it βˆ’ e βˆ’ it ) / 2 i , Z C ( x 3 βˆ’ iy 3 ) dz = 1 8 i Z 2 Ο€ Ο€ ( e 3 it + 3 e it + 3 e βˆ’ it + e βˆ’ 3 it ) e it dt + 1 8 i Z 2 Ο€ Ο€ ( e 3 it βˆ’ 3 e it + 3 e βˆ’ it βˆ’ e βˆ’ 3 it ) e it dt = 1 8 i Z 2 Ο€ Ο€ (2 e 4 it + 6) dt = 3 Ο€ 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 = βˆ’ Ο€x + 2 Ο€ , 1 ≀ x ≀ 2, respectively. Now Z C 1 e z dz = Z 2 e x dx = e 2 βˆ’ 1 Z C 2 e z dz = (1 βˆ’ Ο€i ) Z 1 2 e x +( βˆ’ Ο€x +2 Ο€ ) i dx = (1 βˆ’ Ο€i ) Z 1 2 e (1 βˆ’ Ο€i ) x dx = e 1 βˆ’ Ο€i βˆ’ e 2(1 βˆ’ Ο€i ) = βˆ’ e βˆ’ e 2 . 779 Exercises 18.1 In the second integral we have used the fact that e z has period 2 Ο€i . Thus Z C e z dz = ( e 2 βˆ’ 1) + ( βˆ’ e βˆ’ e 2 ) = βˆ’ 1 βˆ’ 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 βˆ’ cos1 Z C 2 sin z dz = i Z 1 sin(1 + iy ) dy = cos1 βˆ’ cos(1 + i ) . Thus Z C sin z dz = (1 βˆ’ cos1)+(cos1 βˆ’ cos(1+ i )) = 1 βˆ’ cos(1+ i ) = (1 βˆ’ cos1cosh1)+ i sin1sinh1 = 0 . 1663+0 . 9889 i. 13. We have Z C Im( z βˆ’ i ) dz = Z C 1 ( y βˆ’ 1) dz + Z C 2 ( y βˆ’ 1) dz On C 1 , z = e it , 0 ≀ t ≀ Ο€/ 2, dz = ie it dt , y = sin t = ( e it βˆ’ e βˆ’ it ) / 2 i , Z C 1 = ( y βˆ’ 1) dz = 1 2 Z Ο€/ 2 [ e it βˆ’ e βˆ’ it βˆ’ 2 i ] e it dt = 1 2 Z Ο€/ 2 [ e 2 it βˆ’ 1 + 2 ie it ] dt = 1 βˆ’ Ο€ 4 βˆ’ 1 2 i. On C 2 , y = x + 1, βˆ’ 1 ≀ x ≀ 0, z = x + ( x + 1) i , dz = (1 + i ) dx , Z C 2 ( y βˆ’ 1) dz = (1 + i ) Z βˆ’ 1 x dx = 1 2 + 1 2 i. Thus Z C Im( z βˆ’ i ) dz = Β΅ 1 βˆ’ Ο€ 4 βˆ’ 1 2 i ΒΆ + Β΅ 1 2 + 1 2 i ΒΆ = 3 2 βˆ’ Ο€ 4 . 14. Using x = 6cos t , y = 2sin t , Ο€/ 2 ≀ t ≀ 3 Ο€/ 2, z = 6cos t + 2 i sin t , dz = ( βˆ’ 6sin t + 2 i cos t ) dt , Z C dz = βˆ’ 6 Z 3 Ο€/ 2 Ο€/ 2 sin t dt + 2 i Z 3 Ο€/ 2 Ο€/ 2 cos t dt = 2 i ( βˆ’ 2) = βˆ’ 4 i....
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Chapter 18 - 18 Integration in the Complex Plane Exercises...

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