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
omplex Analysis Module: 6: Residue alculus Lecture: 3: ontour integration and applications A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 2 / 18
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
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
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
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 2 + 1. z =1 A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 6 / 18
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
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
Evaluate the integral dθ 3 + 2 sin θ Solution: Substituting z = e iθ, dθ = dz iz, sin θ = (z2 1)/2zi I = = 2 = dθ 3 + 2 sin θ dz z 2 + 3zi 1 z =1 f (z)dz A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 9 / 18
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
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 3 + 2 sin θ = 2πi.( i) = 2π 5 5 A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 11 / 18
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
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
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
Show that Hints: (1 + 2 cos θ) n cos nθ dθ = 2π (3 5) n 3 + 2 cos θ 5 I = = Re = Re 1 i (1 + 2 cos θ) n cos nθ dθ 3 + 2 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
ontinued... Poles of f (z) are α = 3 + 5 2 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 = 3 + 5 2 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
Exercises 1 Evaluate π 2 Evaluate 3 Evalute cos θ 5 + 4 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
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