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Chapter_4 - IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18...

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IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1
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IV.1.1 1 2 3 P L K F KKK FFF Let ° be the boundary of the triangle f 0 < y < 1 ° x; 0 < x < 1 g , with the usual counterclockwise orientation. Evaluate the following integrals (a) R ° Re z dz (b) R ° Im z dz (c) R ° z dz Solution a) We begin do split the contur ° into three parts so that ° = ° 1 + ° 2 + ° 3 . And integrate along each side in the triangle ° = ° 1 + ° 2 + ° 3 We parametrizize ° 1 = [0 ; 1] , thus ° 1 ( t ) = t , for 0 ± t ± 1 , then Re ( t ) = t and dz = dt Z ° 1 Re z dz = Z ° 1 Re ( t ) dz = Z 1 0 t dt = 1 2 : We parametrizize ° 2 = [1 ; i ] , thus ° 2 ( t ) = 1 ° t + it , for 0 ± t ± 1 , then Re (1 ° t + it ) = 1 ° t and dz = ( ° 1 + i ) dt , then Z ° 2 Re z dz = Z 1 0 (1 ° t ) ( ° 1 + i ) dt = ° 1 2 + 1 2 i: We parametrizize ° 3 = [ i; 0] , thus ° 3 ( t ) = i (1 ° t ) , for 0 ± t ± 1 , then, Re ( i (1 ° t )) = 0 and dz = ° i dt , then Z ° 3 Re z dz = Z 1 0 (0) ( ° i ) dt = 0 : 2
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And now we take the parts in the integral together Z ° Re z dz = Z ° 1 Re z dz + Z ° 2 Re z dz + Z ° 3 Re z dz = i 2 b) For details see a) Z ° 1 Im z dz = Z 1 0 0 dt = 0 ; Z ° 2 Im z dz = Z 1 0 t ( ° 1 + i ) dt = ° 1 2 + 1 2 i; Z ° 3 Im z dz = Z 1 0 (1 ° t ) ( ° i ) dt = ° 1 2 i: thus Z ° Im z dz = ° 1 2 : c) For details see a) Z ° 1 z dz = Z 1 0 t dt = 1 2 ; Z ° 2 z dz = Z 1 0 (1 ° t + it ) ( ° 1 + i ) dt = ° 1 Z ° 3 z dz = Z 1 0 ( i (1 ° t )) ( ° i ) dt = 1 2 : thus Z ° z dz = 0 : 3
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IV.1.2 1 2 3 P L K 111 LLL Absolutbeloppet i c ) n ° agot ± ar fel Let ° be the unit circle fj z j = 1 g , with the usual counterclockwise orientation. Evaluate the following integrals, for m = 0 ; ² 1 ; ² 2 ; : : : . (a) R ° z m dz (b) R ° ² z m dz (c) R ° z m j dz j Solution Set z = e , where 0 ± ± ± 2 ² , and thus dz = ie . (a) Z ° z m dz = Z 2 ² 0 e im± ie = i Z 2 ² 0 e i ( m +1) ± = ° e i ( m +1) ± m + 1 ± 2 ² 0 = = 1 m + 1 ² e i 2 ² ( m +1) ° 1 ³ = ´ 0 ; m 6 = ° 1 2 ²i m = ° 1 : (b) Z ° ² z m dz = Z 2 ² 0 e ° im± ie = i Z 2 ² 0 e i (1 ° m ) ± = = ° e i (1 ° m ) ± 1 ° m ± 2 ² 0 = 1 1 ° m ² e i 2 ² ( m ° 1) ° 1 ³ = ´ 0 ; m 6 = 1 2 ²i m = 1 : (c) Z ° z m j dz j = Z 2 ² 0 e im± = ° e im± im ± 2 ² 0 = 1 im ² e i 2 ²m ° 1 ³ = ´ 0 ; m 6 = 0 2 ² m = ° 1 : 4
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IV.1.3 1 2 3 P L K 111 Let ° be the circle fj z j = R g , with the usual counterclockwise ori- entation. Evaluate the following integrals, for m = 0 ; ² 1 ; ² 2 ; : : : . (a) R ° j z m j dz (b) R ° j z m j j dz j (c) R ° ² z m dz Solution Set z = Re , where 0 ± ± ± 2 ² , and thus dz = iRe and j z j = R . (a) Z ° j z m j dz = Z 2 ² 0 µ µ R m e im± µ µ iRe = iR m +1 Z 2 ² 0 e = R m +1 e · 2 ² 0 = R m +1 ² e i 2 ² ° 1 ³ = 0 : (b) Z ° j z m j j dz j = Z 2 ² 0 µ µ R m e im± µ µ µ µ iRe µ µ = R m +1 Z 2 ² 0 = R m +1 [ ± ] 2 ² 0 = R m +1 (2 ² ° 0) = 2 ²R m +1 : (c) Z ° ² z m dz = Z 2 ² 0 R m e ° im± iRe = iR m +1 Z 2 ² 0 e i (1 ° m ) ± = R m +1 ° e i (1 ° m ) ± 1 ° m ± 2 ² 0 = = R m +1 1 ° m ² e i 2 ² (1 ° m ) ° 1 ³ = ´ 0 ; m 6 = 1 ; 2 ²iR m +1 ; m = 1 ; = ´ 0 ; m 6 = 1 ; 2 ²iR 2 ; m = 1 : 5
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IV.1.4 1 2 3 P L K LLL Show that if D is a bounded domain with smooth boundary, then Z @D ² z dz = 2 i Area ( D ) : Solution Substitute ² z = x ° iy , dz = dx + idy , and apply Green°s theorem.
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