Chapter_4

Chapter_4 - 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 IV.1.1 1 2 3 P L K F KKK FFF Let& be the boundary of the

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 IV.1.1 1 2 3 P L K F KKK FFF Let & be the boundary of the triangle f < y < 1 & x; < 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 ¡ t ¡ 1 , then Re ( t ) = t and dz = dt Z & 1 Re z dz = Z & 1 Re ( t ) dz = Z 1 tdt = 1 2 : We parametrizize & 2 = [1 ;i ] , thus & 2 ( t ) = 1 & t + it , for ¡ t ¡ 1 , then Re (1 & t + it ) = 1 & t and dz = ( & 1 + i ) dt , then Z & 2 Re z dz = Z 1 (1 & t ) ( & 1 + i ) dt = & 1 2 + 1 2 i: We parametrizize & 3 = [ i; 0] , thus & 3 ( t ) = i (1 & t ) , for ¡ t ¡ 1 , then, Re ( i (1 & t )) = 0 and dz = & idt , then Z & 3 Re z dz = Z 1 (0) ( & i ) dt = 0 : 2 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 dt = 0 ; Z & 2 Im z dz = Z 1 t ( & 1 + i ) dt = & 1 2 + 1 2 i; Z & 3 Im z dz = Z 1 (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 tdt = 1 2 ; Z & 2 z dz = Z 1 (1 & t + it ) ( & 1 + i ) dt = & 1 Z & 3 z dz = Z 1 ( i (1 & t )) ( & i ) dt = 1 2 : thus Z & z dz = 0 : 3 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 ; & 1 ; & 2 ;::: . (a) R & z m dz (b) R & ¢ z m dz (c) R & z m j dz j Solution Set z = e i¡ , where ¡ ¡ ¡ 2 ¢ , and thus dz = ie i¡ d¡ . (a) Z & z m dz = Z 2 ¢ e im¡ ie i¡ d¡ = i Z 2 ¢ e i ( m +1) ¡ d¡ = & e i ( m +1) ¡ m + 1 ¡ 2 ¢ = = 1 m + 1 ¢ e i 2 ¢ ( m +1) ¢ 1 £ = ¤ ; m 6 = ¢ 1 2 ¢i m = ¢ 1 : (b) Z & ¢ z m dz = Z 2 ¢ e & im¡ ie i¡ d¡ = i Z 2 ¢ e i (1 & m ) ¡ d¡ = = & e i (1 & m ) ¡ 1 ¢ m ¡ 2 ¢ = 1 1 ¢ m ¢ e i 2 ¢ ( m & 1) ¢ 1 £ = ¤ ; m 6 = 1 2 ¢i m = 1 : (c) Z & z m j dz j = Z 2 ¢ e im¡ d¡ = & e im¡ im ¡ 2 ¢ = 1 im ¢ e i 2 ¢m ¢ 1 £ = ¤ ; m 6 = 0 2 ¢ m = ¢ 1 : 4 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 ; & 1 ; & 2 ;::: . (a) R & j z m j dz (b) R & j z m j j dz j (c) R & & z m dz Solution Set z = Re i¡ , where ¡ ¡ ¡ 2 ¢ , and thus dz = iRe i¡ d¡ and j z j = R . (a) Z & j z m j dz = Z 2 ¢ & & R m e im¡ & & iRe i¡ d¡ = iR m +1 Z 2 ¢ e i¡ d¡ = R m +1 ¡ e i¡ ¢ 2 ¢ = R m +1 £ e i 2 ¢ ¢ 1 ¤ = 0 : (b) Z & j z m jj dz j = Z 2 ¢ & & R m e im¡ & & & & iRe i¡ & & d¡ = R m +1 Z 2 ¢ d¡ = R m +1 [ ¡ ] 2 ¢ = R m +1 (2 ¢ ¢ 0) = 2 ¢R m +1 : (c) Z & & z m dz = Z 2 ¢ R m e & im¡ iRe i¡ d¡ = iR m +1 Z 2 ¢ e i (1 & m ) ¡ d¡ = R m +1 ¥ e i (1 & m ) ¡ 1 ¢...
View Full Document

This note was uploaded on 11/13/2010 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.

Page1 / 54

Chapter_4 - 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 IV.1.1 1 2 3 P L K F KKK FFF Let& be the boundary of the

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online