NPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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1 NPTEL web course on omplex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 1 / 18
2 omplex Analysis Module: 6: Residue alculus Lecture: 3: ontour integration and applications A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 2 / 18
3 Evaluation of definite integrals using Residues Here Residue theorem is used to evaluate various type of contour integral. The basic method will always be: set up the integral of a suitable function f around a suitable simple-closed contour Γ: identify the poles of f which lie inside Γ; calculate the residues of f at these poles. A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 3 / 18
4 Evaluation of definite integrals using Residues Type I The integrals of the form R(cos θ, sin θ) dθ where the integrand is a rational function of cos θ and sin θ. A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 4 / 18
5 Procedure We integrate along a unit circle and make the substitute z = e iθ.then, cos θ = 1 ( z + 1 ), sin θ = 1 ( z 1 ) 2 z 2i z also dz = ie iθ dθ; dθ = dz iz. Thus the integral is transformed to line integral R(cos θ, sin θ) dθ = 1 i z =1 ( ( 1 R z + 1 ), 1 ( z 1 )) dz 2 2 2i z z. The contour integral given has to be evaluated using the residue theorem. A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 5 / 18
6 Evaluate the integral where a > 1. Solution: Since cos θ is an even function, π dθ a + cos θ dθ a + cos θ = 1 dθ 2 a + cos θ. Substituting the values z = e iθ, dθ = dz in the integral, we get the iz modified integral 1 dz i 2az + z z =1 A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 6 / 18
7 continued... 1 The poles of f (z) = z 2 + 2az + 1 are given by α = a + a 2 1 and β = a a 2 1. We find that α < 1 and β > 1. learly α lies inside : z = 1 and β lies outside : z = 1. 1 The Residue of f at α is given by 2 a 2 1. A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 7 / 18
8 continued... From the calculus of residues we have f (z) dz = 2πi jεs Res(f, α j ). The solution of the integral is π dθ a + cos θ =2πi. 1 [ ] 1 i a 2 1 π = a 2 1. A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 8 / 18
9 Evaluate the integral dθ sin θ Solution: Substituting z = e iθ, dθ = dz iz, sin θ = (z2 1)/2zi I = = 2 = dθ sin θ dz z 2 + 3zi 1 z =1 f (z)dz A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 9 / 18
10 ontinued... The zeros of f (z) are a = ( 3 + 5)i and b = ( 3 5)i 2 2 learly a < 1 and b > 1. So, a lies inside : z = 1 and b lies outside : z = 1. The residue of f at z = a is given by i. 5 A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 1 / 18
11 ontinued... From the calculus of residues we have f (z)dz = 2πi Res(f, α j ) j S The solution of the integral is dθ sin θ = 2πi.( i) = 2π 5 5 A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 11 / 18
12 Prove that e cos θ. cos(sin θ nθ)dθ = 2π n!, n N Proof. := {z : z = 1} I = = Re = Re e cos θ. cos(sin θ nθ)dθ e cos θ.e i(sin θ nθ) dθ exp[cos θ + i(sin θ nθ)]dθ A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 12 / 18
13 ontinued... = Re exp(e iθ inθ)dθ inθ dz = Re exp(z)e iz, z = eiθ = Re 1 i = Re 1 i e z dz zn+1 f (z)dz A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 13 / 18
14 ontinued... f (z) has pole at z = of order n + 1 Residue of f (z) at z = is 1 n! By auchy s residue theorem, f (z)dz = 2πi n! I = Re 1 i 2πi n! = 2π n! A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 14 / 18
15 Show that Hints: (1 + 2 cos θ) n cos nθ dθ = 2π (3 5) n cos θ 5 I = = Re = Re 1 i (1 + 2 cos θ) n cos nθ dθ cos θ (1 + e iθ + e iθ ) n e inθ 3 + e iθ + e iθ dθ (z 2 + z + 1) n z 2 + 3z + 1 dz = Re1 i f (z)dz, z = e iθ A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 15 / 18
16 ontinued... Poles of f (z) are α = α < 1 and β > 1 Residue of f (z) at z = α is ( 3 + 5) n 5 By auchy s residue theorem, I = Re 1 i and β = f (z) = 2πi(3 5) n 5 2πi (3 5) n = 2π (3 5) n 5 5 A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 16 / 18
17 Exercises 1 Evaluate π 2 Evaluate 3 Evalute cos θ cos 2θ dθ dθ cos 2 θ + 4 sin 2 θ sin nθ 1 + 2a cos θ + a 2, cos nθ 1 + 2a cos θ + a 2 A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 17 / 18
18 ontinued... Hints: Take I = = 1 ai e inθ 1 + 2a cos θ + a 2 = z n z 2 + az + z a + 1 dz, (e iθ ) n 1 + 2a cos θ + a 2 for z = e iθ = 2π( 1)n a n 1 a 2, using auchy s residue theorem Equating real and imaginary parts, we get the answers. A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 18 / 18
NPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 28 Complex Analysis Module: 6:
More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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More informationNPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India
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