AEM_3e_Chapter18

AEM_3e_Chapter18 - 18 1 C Integration in the Complex Plane...

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18 18 Integration in the Complex Plane EXERCISES 18.1 Contour Integrals 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 0 [ t 3( t 2 +2) i ]( 1+2 ti ) dt = Z 2 0 (6 t 3 +13 t ) dt + i Z 2 0 ( 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 0 ( 15 t 4 +4 t 3 +3 t 2 2 t ) dt + i Z 1 0 ( 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 0 cos t ( sin t + i cos t ) dt = Z 2 π 0 sin t cos tdt + i Z 2 π 0 cos 2 tdt = 1 2 Z 2 π 0 sin2 tdt + 1 2 i Z 2 π 0 (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 0 [ 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 0 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. 877
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18.1 Contour Integrals 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 0 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 . In the second integral we have used the fact that e z has period 2 πi .Thu s Z C e z dz =( e 2 1)+( e e 2 )= 1 e. 12. Z C sin zdz = Z C 1 sin zdz + Z C 2 sin zdz where C 1 and C 2 are the line segments y =0 ,0 x 1, and x =1 , 0 y 1, respectively. Now Z C 1 sin zdz = Z 1 0 sin xdx =1 cos1 Z C 2 sin zdz = i Z 1 0 sin(1 + iy ) dy = cos1 cos(1 + i ) . Thus Z C sin zdz =(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 0 [ e it e it 2 i ] e it dt = 1 2 Z π/ 2 0 [ e 2 it 1+2 ie it ] dt =1 π 4
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AEM_3e_Chapter18 - 18 1 C Integration in the Complex Plane...

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