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Unformatted text preview: Complex Analysis Homework 7, Thanks to Da Zhang for providing the texfile 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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 Summer '08
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 Konrad Zuse, dz, Resf, ΓR

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