AEM_3e_Chapter_18

AEM_3e_Chapter_18 - 18 18 3 C Integration in the Complex...

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18 18 Integration in the Complex Plane EXERCISES 18.1 Contour Integrals 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 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 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. 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. 15. We have I ˇ C ze z dz = Z C 1 ze z dz + Z C 2 ze z dz + Z C 3 ze z dz + Z C 4 ze z dz On C 1 , y =0,0 x 1, z = x , dz = dx , Z C 1 ze z dz = Z 1 0 xe x dx = xe x e x ¯ ¯ ¯ 1 0 =1 . On C 2 , x =1,0 y 1, z =1+ iy , dz = idy , Z C 2 ze z dz = i Z 1 0 (1 + iy ) e 1+ iy dy = ie i +1 . On C 3 , y =1,0 x 1, z = x + i , dz = dx , Z C 3 ze z dz = Z 0 1 ( x + i ) e x + i dx =( i 1) e i ie 1+ i
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18.1 Contour Integrals On C 4 , x =0,0 y 1, z = iy , dz = idy , Z C 4 ze z dz = Z 0 1 ye iy dy =(1 i ) e i 1 . Thus I ˇ C ze z dz =1+ ie i +1 +( i 1) e i ie 1+ i +(1 i ) e i 1=0 . 18. We have I ˇ C (2 z 1) dz = Z C 1 (2 z 1) dz + Z C 2 (2 z 1) dz + Z C 3 (2 z 1) dz On C 1 , y =0,0 x 1, z = x , dz = dx , Z C 1 (2 z 1) dz = Z 1 0 (2 x 1) dx =0 . On C 2 , x =1,0 y 1, z =1+ iy , dz = idy , Z C 2 (2 z 1) dz = 2 Z 1 0 ydy + i Z 1 0 dy = 1+ i. On C 3 , y = x , z = x + ix , dz =(1+ i ) dx , Z C 3 (2 z 1) dz =(1+ i ) Z 0 1 (2 x 1+2 ix ) dx =1 i. Thus I ˇ C (2 z 1) dz =0 1+ i +1 i =0 . 21. On C , y = x +1,0 x 1, z = x +( x +1) i , dz =(1 i ) dx , Z C ( z 2 z +2) dz =(1 i ) Z 1 0 [ x 2 (1 x ) 2 x +2+(3 x 2 x 2 1) i ] dx = 4 3 5 3 i. 24.
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AEM_3e_Chapter_18 - 18 18 3 C Integration in the Complex...

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