HW7s - Complex Analysis Homework 7, Thanks to Da Zhang for...

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Unformatted text preview: Complex Analysis Homework 7, Thanks to Da Zhang for providing the tex-file November 7, 2011 Exercises in Chapter 4 of the textbook. 34. (b). 1 2 πi H ∂D (0 , 5) f ( z ) dz where f ( z ) = e z ( z +1) sin z . Solution. Inside D (0 , 5), f ( z ) = e z ( z +1) sin z has four singularities(poles)- π ,- 1, 0, π ; moreover, all of them has order of 1, So calculating the residues gives us: Res f (- π ) = lim z →- π ( z + π ) e z ( z + 1)sin z =- e- π 1- π Res f (- 1) = lim z →- 1 ( z + 1) e z ( z + 1)sin z = e- 1 sin(- 1) Res f (0) = lim z → ze z ( z + 1)sin z = 1 Res f ( π ) = lim z → π ( z- π ) e z ( z + 1)sin z =- e π 1 + π Also, the index of the curve ∂D (0 , 5) w.r.t each of the singularities is 1, so using 1 the residue formula, we calculate the integral as 1 2 πi I ∂D (0 , 5) f ( z ) dz = Res f (- π ) + Res f (- 1) + Res f (0) + Res f ( π ) =- e- π 1- π + e- 1 sin(- 1) + 1- e π 1 + π (d). 1 2 πi H ∂D (0 , 5) f ( z ) dz where f ( z ) = e z z ( z +1)( z +2) and γ is the negatively (clockwise) oriented triangle with vertices 1 ± i and- 3. Solution. It is obvious that f ( z ) has three poles:- 2,- 1, 0, and their orders are all one. 2 Hence, calculating the residues gives us: Res f (- 2) = lim z →- 2 ( z + 2) e z z ( z + 1)( z + 2) = e- 2 2 Res f (- 1) = lim z →- 1 ( z + 1) e z z ( z + 1)( z + 2) =- e- 1 Res f (0) = lim z → ze z z ( z + 1)( z + 2) = 1 2 Also, the index of the triangular curve w.r.t each of the poles is- 1, so using the residue formula, we get 1 2 πi I ∂D (0 , 5) f ( z ) dz = Res f (- 2) + Res f (- 1) + Res f (0) =- e- 2 2 + e- 1- 1 2 (h). 1 2 πi H ∂D (0 , 5) f ( z ) dz where f ( z ) = e iz (sin z )(cos z ) and γ is the positively (coun- terclockwise) oriented quadrilateral with vertices ± 5 i , ± 10. 3 Solution. Inside this quadrilateral, we find that f ( z ) has 13 poles: ± 3 π , ± 5 π 2 , ± 2 π , ± 3 π 2 , ± π , ± π 2 , 0. Moreover, they are all poles of order 1. So the residues can be calculated as: Res f (0) = 1 Res f (- π 2 ) = i Res f ( π 2 ) =- i Res f (- π ) =- 1 Res f ( π ) =- 1 Res f (- 3 π 2 ) =- i Res f ( 3 π 2 ) = i Res f (- 2 π ) = Res f (- 2 π ) = 1 Res f (- 5 π 2 ) = i Res f ( 5 π 2 ) =- i Res f (- 3 π ) =- 1 Res f (3 π ) =- 1 Also, the index of the quadrilateral w.r.t each of the poles is one, so applying the residue theorem gives us: 1 2 πi I ∂D (0 , 5) f ( z ) dz = 6 X k =- 6 Res f ( kπ 2 ) =- 1 2 47. Compute Z + ∞-∞ cos x 1 + x 4 dx Solution. Consider the following holomorphic function and contour γ R : f ( z ) = e iz 1 + z 4 4 We know that f ( z ) has two poles inside γ R :...
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This note was uploaded on 02/21/2012 for the course MATH 4377 taught by Professor Staff during the Summer '08 term at University of Houston.

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HW7s - Complex Analysis Homework 7, Thanks to Da Zhang for...

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