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1(z+i)5at-i3.z12(z-1)10at 1Solution.1. We see that the functionf(z) =z+ 1zsin (1/z)has a pole of order 2 atz= 0. We needto prove thatlimz→0z2f(z) = 0Indeed, we havez2z+ 1zsin (1/z)≤ |z(z+ 1)|We know thatlimz→0z(z+ 1) = 0, so we obtain the proof.2. We see that the function has a pole of order 6 atz=-i. Indeed, we havelimz→-i(z+i)61(z+i)5= limz→-i(z+i) = 03. In a similar fashion, the function has a pole of order 11 atz= 1. We see this as follows178
6 RESIDUES AND THEIR USE IN INTEGRATIONlimz→1(z-1)11z12(z-1)10= limz→1z12(z-1) = 0Exercise 6.9.(E. 27, E. 31 P. 358)Use residues to evaluate the following integrals. Use the principal branch of multivalued func-tions.1.‰dzsinzaround|z-6|= 42.‰e1/zz2-1dzaround|z-1|= 3/2Solution.179
6 RESIDUES AND THEIR USE IN INTEGRATION6.4 EVALUATION OF REAL INTEGRALS WITH RESIDUECALCULUS IThe reader can see exercises 3, 4, 7, 8 for examples of residue calculus I.Exercise 6.10.(E. 3, E. 4 P. 364)Using residue calculus, establish the following identities.1.ˆπ-πdθa+bcosθ=2π√a2-b2fora > b≥0.2.ˆ3π2-π2cosθa+bcosθdθ=2πb1-a√a2-b2fora > b >0.Solution.1. We putt=π+θand see thatˆπ-πdθa+bcosθ=ˆ2π0dθa-bcosθThe next is the denominator is definited because ofa > b≥0. Then we apply the methodof evaluation of real integrals with residue calculus 1. We getz=eiθ, so thatdθ=dzizcosθ=z+z-12Then the integral becomesˆ2π0dθa-bcosθ=‰|z|=12dz2iaz-ibz2-ibWe need to face with the equationibz2-2iaz+ib= 0and have 2 roots easily.z=a±√a2-b2bBut the only point,z=a-√a2-b2b, is in the unit circle, the reader can verify themselves.Now we only compute the residue at this point.ˆ2π02dθ2iaz-ibz2-ib=4πiRes12iaz-ibz2-ib;a-√a2-b2b!=2π√a2-b2Note that, we only get the result in the casea > b >0, but do not worry, it is very easyto get the proof for the casesa > b= 0.180
6 RESIDUES AND THEIR USE IN INTEGRATION2. Like the exercise above, we putt=θ+π2and see thatˆ3π2-π2cosθa+bcosθdθ=ˆ2π0sinθa+bsinθdθThe next is the denominator is definited because ofa > b >0. Then we apply the methodof evaluation of real integrals with residue calculus 1. We getz=eiθ, so thatdθ=dzizsinθ=z-z-12iThen the integral becomesˆ2π0sinθa+bsinθdθ=‰|z|=1(z2-1)dziz(bz2+ 2iaz-b)We need to face with the equationbz2+ 2iaz-b= 0and have 2 roots easily.z=-ia±i√a2-b2bBut the only point,z=-ia+i√a2-b2b, is in the unit circle, the reader can verify them-selves.Now we only compute the residue at this point and the pointz= 0.ˆ2π0sinθa+bsinθdθ=2πi"Resz2-1bz2+ 2iaz-b;-ia+i√a2-b2b!+Resz2-1bz2+ 2iaz-b; 0#=2πb1-a√a2-b2Exercise 6.11.(E. 7, E. 8, E. 9 P. 364)Using residue calculus, establish the following identities.1.ˆ2π0dθ(a+bsinθ)2=2πa√a2-b23fora > b≥0.2.ˆ2π0dθa+ sin2θ=2πpa(a+ 1)fora >0.