Chapter_7

Chapter_7 - VII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17...

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Unformatted text preview: VII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 X X X X 2 X X X X X X X X X X X X 3 X X X X X X X 4 X X X X X X X X X X 5 X X X X X X 6 X X X X X X X 7 X X X 8 X 1 VII.1.1 Evaluate the following residues. (a) Res & 1 z 2 +4 ; 2 i ¡ (d) Res & sin z z 2 ; ¡ (g) Res h z Log z ; 1 i (b) Res & 1 z 2 +4 ; & 2 i ¡ (e) Res & cos z z 2 ; ¡ (h) Res & e z z 5 ; ¡ (c) Res & 1 z 5 & 1 ; 1 ¡ (f) Res [cot z; 0] (i) Res & z n +1 z n & 1 ; w k ¡ Solution a) By rule 4, Res ¢ 1 z 2 + 4 ; 2 i £ = 1 2 z ¤ ¤ ¤ ¤ z =2 i = 1 4 i = & i 4 : b) By rule 4, Res ¢ 1 z 2 + 4 ; & 2 i £ = 1 2 z ¤ ¤ ¤ ¤ z = & 2 i = & 1 4 i = i 4 : c) By rule 4, Res ¢ 1 z 5 & 1 ; 1 £ = 1 5 z 4 ¤ ¤ ¤ ¤ z =1 = 1 5 : d) By rule 1, Res ¢ sin z z 2 ; £ = lim z ! sin z z = 1 : e) By rule 2, Res h cos z z 2 ; i = lim z ! d dz cos z = & sin z j z =0 = 0 : f) By rule 3, Res [cot z; 0] = Res h cos z sin z ; i = cos z cos z ¤ ¤ ¤ z =0 = 1 : 2 g) By rule 3, Res & z Log z ; 1 ¡ = z 1 =z ¢ ¢ ¢ ¢ z =1 = 1 : h) By Laurent expansion e z z 5 = 1 + z 1! + z 2 2! + z 3 3! + z 4 4! + o ( z 5 ) z 5 = 1 z 5 + 1 1! z 4 + 1 2! z 3 + 1 3! z 2 + 1 4! z + o (1) ; hence Res & e z z 5 ; ¡ = 1 4! = 1 24 : i) We have poles of the form w k = e 2 &ik=n , where k = 0 ; 1 ; 2 ; : : : n & 1 , remark that w n k = 1 , by rule 3, Res & z n + 1 z n & 1 ; w k ¡ = z n + 1 nz n & 1 ¢ ¢ ¢ ¢ z = w k = w n k + 1 nw n & 1 k = 2 nw n & 1 k = 2 w k n = 2 e 2 &ik=n n : 3 VII.1.2 Calculate the residue at each singularity in the complex plane of the following functions. (a) e 1 =z (b) tan z (c) z ( z 2 +1) 2 (d) 1 z 2 + z Solution (a) The function e 1 =z has a singularity at z = 0 , by Laurent expansion e 1 =z = 1 + 1 z + 1 2! z 2 + : : : : Thus by de&nition Res & e 1 =z ; ¡ = 1 : (b) The function tan z = sin z cos z has isolated singularities at z = & 2 + &n , &1 < n < 1 , they are simple poles. By rule 3, Res [tan z; &= 2 + n& ] = Res ¢ sin z cos z ; &= 2 + n& £ = sin z & sin z ¤ ¤ ¤ ¤ z = &= 2+ n& = & 1 : (c) The function z ( z 2 +1) 2 has isolated singularities at z = ¡ i , they are double poles. By rule 2, Res ¢ z ( z 2 + 1) 2 ; i £ = d dz z ( z + i ) 2 ¤ ¤ ¤ ¤ z = i = ( z + i ) 2 & 2 z ( z + i ) ( z + i ) 4 ¤ ¤ ¤ ¤ ¤ z = i = 0 ; Res ¢ z ( z 2 + 1) 2 ; & i £ = d dz z ( z & i ) 2 ¤ ¤ ¤ ¤ z = & i = ( z & i ) 2 & 2 z ( z & i ) ( z & i ) 4 ¤ ¤ ¤ ¤ ¤ z = & i = 0 : (d) 4 The function 1 z ( z +1) has isolated singularities at z = 0 and z = & 1 they are simple poles. By rule 1, Res & 1 z ( z + 1) ; ¡ = lim z ! 1 z + 1 = 1 z + 1 ¢ ¢ ¢ ¢ z =0 = 1 ; Res & 1 z ( z + 1) ; & 1 ¡ = lim z !& 1 1 z = 1 z ¢ ¢ ¢ ¢ z = & 1 = & 1 : 5 VII.1.3 Evaluate the following integrals using residue theorem....
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This note was uploaded on 11/13/2010 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.

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Chapter_7 - VII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17...

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