Complex Variables...Review Problems (Residue Calculus Comments)...Fall Initial Draft

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1 Complex Variables Review Problems Residue Calculus Comments) Fall 22 Initial Draft ) Show that the singular point of fz) is a pole; determine its order m and its residue B: a) e 2z )/z 4, b) e 2z /z ) 2. Comments: a) The order of the pole is m = 3 and the residue is B = 4/3. To verify this, we use the Taylor expansion e 2z = + 2z) + 2! 2z)2 + 3! 2z)3 + to obtain e 2z z 4 = = k= k= so the negative powers of the expansion are )/ k! 2z)k z 4 k! 2k z k 4 2 z ! z ! z. Hence the order of the pole at z = is 3 and the residue given by the coefficient of /z is 2 3 /3! = 4/3. b) The order of the pole is m = 2 and the residue is B = 2e 2. We use the Taylor expansion of e 2z about z = : e 2z = k= f k) ) z ) k = k! since D k e 2k ) = 2 k e 2k. Then we find that so the negative powers are e 2z z ) 2 = k= k= 2 k e 2 z ) k 2 k! e 2 z ) 2 + 2e 2 z ). 2 k e 2 z ) k k! We conclude that is a pole of order 2 and its residue is 2e 2. 2) Evaluate the following integrals around the circle z = 3: a) e z /z 2, b) e z /z ) 2, c) z 2 e /z. Comments: These integrals can all be found using the Residue Theorem. a) Let fz) = e z /z 2 which has a unique pole at z = of order 2. By the Residue Theorem, we have z =3 e z z 2 dz = 2πiResfz); ) = 2πi lim z d dz z 2 e z z 2 ) = 2πie = 2πi. b) Let fz) = e z /z ) 2. Then fz) has a unique pole at z = which has order 2. By

2 applying the Residue Theorem, we find that z =3 e z dz = 2πiResfz); ) z ) 2 d = 2πi lim z dz = 2πie = 2πi z ) 2 e z z ) 2 ) c) Let fz) = z 2 e /z which has a unique singularity at z = which is essential. Note that fz) = z 2 + z + 2! z 2 + 3! z 3 + ) 4! z 4 + = z 2 + z + 2! + 3! z + 4! z 2 + which shows that Resfz); ) = /6. By the Residue Theorem, we must have z =3 z 3 e /z dz = 2πiResfz); ) = 2πi 6 = πi 3. Review: Identifying the order of a pole if fz) = pz)/qz). Assume that pa), qa) = but q a). Then write pz) qz) = pz) q a)z a) + 2! q a)z a) 2 + = z a pz) q a) + 2! q a)z a) +. Note that both pz) and q a) + 2! q a)z a) + are analytic functions around z = a and are both nonzero at z = a. Hence the order of the pole is indeed. 3) Evaluate the following residues: a) fz) = z /4 /z i) where z /4 principal branch); Resfz); i). b) fz) = Logz)/z 2 + ) 2 principal branch); Resfz); i). c) fz) = e z / sin z; Resfz); π). Comments: a) f has a simple pole at z = i so we find that Resfz); i) = lim z i z i) z/4 z i = i/4 = cosπ/8) + i sinπ/8) 2

3 since i = e π/2. b) fz) has a pole of order 2 at so the residue is given as ) d lim z i) 2 Log z d Log z z i dz z 2 + ) 2 = lim z i dz z + i) 2 z + i) 2 /z 2z + i)log z = lim z i z + i) 4 ) 2Log z = lim z i z + i) 3 zz + i) 2 = 2Log i 2i) 3 i2i) 2 = 8 π + 4 i c) fz) has a simple pole at z = π since sin z has only simple zeros since its derivative is nonzero there. Hence the residue is given as Resfz); π) = ez cos z = e π. z=π 4) Let C be the circle z = 2. Evaluate the following integrals: a) tan z dz, b) C C sin 2z dz, cos πz C zz 2 + ) dz Comments: a) Let fz) = tan z. fz) has simple poles at integer multiples of π/2; in particular, the two poles at ±π/2 are lie inside the circle z = 2. Now Resfz); π/2) = and Resfz); π/2) =. By the Residue Theorem, we have tanz) dz = 2πi 2) = 4πi. C b) Let fz) = / sin2z). Then fz) has three simple poles inside the circle z = 2 at and ±π/2. We find that Resfz); ) = /2, Resfz); π/2) = /2, and Resfz); π/2) = /2. By the Residue Theorem, we find that dz = 2πi sin 2z 2 = πi. c) Let fz) = cos πz zz 2 +) C which has three simple poles at and ±i. We find that Resfz); ) =, Resfz); i) = e π /4 e π /4, and Resfz); i) = e π /4 e π /4. By the Residue Theorem, we have C cos πz zz 2 + ) dz = 2πi + e π /4 e π /4) + e π /4 e π /4) ). Review: applying the Residue Theorem to evaluate integrals of the form I = fx) when f can be extended as an analytic function in a domain that includes the real line and the upper half plane except for finitely many poles with positive imaginary part. Let C R denote the semicircle z = R with Iz ; let Γ R be the closed semicircle consisting of C R and the line segment [ R, R] oriented in the positive sense. Then ) I = lim R R fx) + R 3 C R fz) dz

4 provided the limit lim fz) dz =. R C R In many cases, we can show that this limit is by using the ML-inequality while in more subtle cases we need to use Jordan s Lemma that we describe below in Problem # 7. { sin x Note that for integrals of the form gx), the corresponding analytic function used is gz)e iz because the contribution of the semicircle in the upper half plane can cos x usually be shown goes to in the limit. 5) Use residues to compute a) x 2 + ) 2, b) d) x 2 + 2x + 3, e) Comments: a) π/4 b) π/2 2). x 4 +, c) x 2 + )x 2 + 2x + 3). x 4 x 2 + 9)x 2 + 4) 2, z c) Let fz) = 4 whose poles that lie in the upper half plane are 3i of order and z 2 +9)z 2 +4) 2 2i of order 2. We have Resfz); 3i) = 27i/5 and Resfz); 2i) = 23i/5. The Residue Theorem will give x 4 4π x 2 + 9)x 2 = 2πi 2i/25) = + 4) 2 25 provided we can show that lim fz) dz = R C R where C R is the semicircle z = R with Iz. To check this, we use the ML-inequality. On C R we have z 4 z 2 + 9)z 2 + 4) 2 R 4 R 2 9)R 2 4) 2, R > 3. Hence we conclude that lim z R CR 4 z 2 + 9)z 2 + 4) 2 dz lim πr R 4 R R 2 9)R 2 4) 2 =. d) Let fz) = /z 2 + 2z + 3) has one pole + i 2 in the upper half plane which is simple. The residue of fz) there is i 2/4. The Residue Theorem yields i 2 x 2 + )x 2 = 2πi = π 2 + 2x + 3) 4 2 provided we can show that lim fz) dz = R C R where C R is the semicircle z = R with Iz. To check this, we use the ML-inequality. On C R we have z 2 + 2z + 3 R + ) R 2, R >. 4

5 Hence we conclude that lim R z 2 + 2z + 3 dz lim πr R R 2 =. C R e) Let fz) = z 2 +)z 2 +2z+3). fz) has two poles i and + i 2 in the upper half plane which are both simple. Then Resfz); i) = /8 i/8 and Resfz); + i 2) = /8. The Residue Theorem shows that x 2 + )x 2 + 2x + 3) = 2πi /8 i/8 + /8) = π 4 provided we can show that lim fz) dz = R C R where C R is the semicircle z = R with Iz. To check this, we use the ML-inequality. On C R we have z 2 + )z 2 + 2z + 3) R 2 + ) [R + ) 2 + 2] R 4, R >. Hence we conclude that lim R z 2 + )z 2 + 2z + 3) lim πr R R 4 =. C R 6) Use residues to compute cos x a) x 2 + a 2 )x 2 + b 2 ) = π beb + ae a )e a b a 2 + b 2, < b < a; b) )ab < a; c) e) f) cos ax x 2 + b 2 ) πe ab =, < a, < b; d) 2 b x 3 sin ax x 2 + 4) 2 = πa )e 2a, < a; x 3 sin x x 2 + )x 2 + 9) = 9 64 πe e π. x sin 2x x = 3 πe 2 ; 7) Use residues to find the principal values of the integrals below: sin x a) x 2 + 4x + 5, x + ) cos x x 2 + 4x + 5, cos x x + a) 2, < b. + b2 cos ax x 2 + = πe a, Comments: a) Write fz) = e iz /z 2 + 4z + 5) which has simple poles at 2 ± i. Futher, Resfz); 2 + i) = e 2i /[2 2 + i) + 4] = 2 e 2i i. We want to justify that [ sin x x 2 + 4x + 5 = I 2πi ]) 2 e 2i i = Iπe 2i ) = πe sin 2) To do this, let C R denote the semicircle z = R with Iz. We consider e iz z 2 + 4z + 5 dz πr R 2) 2 as R z =R 5

6 since e iz = e ireiθ = e R sin θ since θ π. Further, write z 2 +4z +5 as z +2) + so by repeated uses of the reversed triangle inequality we get Re iθ + 2) 2 + Re iθ + 2) 2 Re iθ Re iθ 2 2 R 2) 2. b) Let fz) = z+)eiz. It has a unique pole in the upper half plane at 2 + i with residue z 2 +4z+5 + i)e 2i /[2i] = 2 + i)e 2i i = 2 + i)e 2i. We will argue that x + ) cos x x 2 + 4x + 5 = R We need to use Jordan s Lemma: π 2πi [ + i)e 2i 2 ]) = πe R + i)cos2) i sin2))) = πe sin2) + cos2)) e R sin θ dθ π R e R ) < π R. Let C R be the portion of the circle z = R with Iz. So C R is parametrized as z = Re iθ, dz = ire iθ dθ with θ π. Then we have the bounds z + )e CR iz z 2 + 4z + 5 dz θ = Re iθ + )e ireiθ Re iθ ) 2 + 4Re iθ + 5 ireiθ dθ π R + ) e ireiθ R 2) 2 + R dθ = R2 + R R 2) 2 < R2 + R R 2) 2 π e R sin θ dθ π as R. R Review: let Rs, t) be a rational function in s and t. Then integrals of the form 2π Rcos t, sin t) dt can be evaluated using the residue theorem by making the substitutions cos t = z + z 2 to obtain the contour integral z =, sin t = z z, dt = dz 2i iz z + z R, z ) z dz 2 2i iz. 6

7 8) Use residue to evaluate 2π a) sin θ dθ = 2π 3, 2π + sin 2 θ dθ = π 2, 2π cos 2 3θ 5 4 cos 2θ dθ = 3π 8. Comments: a) We make the substitutions indicated in the above review: 2π sin θ dθ = z = = 2 dz 5 + 4[z z )/2i)] iz z = z = z = 5z 2iz 2 ) dz 2iz + 2i)z + i/2) dz z + 2i)z + i/2) dz Let fz) = /[z + 2i)z + i/2))] which has only one pole i/2 that lies inside the unit circle. The corresponding residue equals / i/2 + 2i) = 2 3i. We conclude that z = dz 5 + 4[z z )/2i)] iz = 2 z = z + 2i)z + i/2) dz = 2πi Resfz); i/2) 2 = πi 23 ) i = 2 3 π. b) We make the indicated substitution outlined in the above review: 2π + sin 2 θ dθ = z = = 4i = 4i dz + [z z )/2i)] 2 iz z = z = z = z = z = = 4i z = dz 4 z z ) 2 z dz 4 z2 2 + z 2 ) z z 4 z3 2z + z ) dz 4z z 3 2z + z ) dz z 3 dz + 6z z z z 4 6z 2 + dz. Let qz) = z 4 6z 2 +, the denominator polynomial. We can find its roots by the quadratic formula applied to w 2 6w + which has roots w = 3 ± 2 2. Since w 2 = z, the roots of qz) are two sets of values given as: ± and z ± = ± The latter two roots z ± lie inside the unit circle while the other do not so they can be discarded below. 7

8 By the Residue Theorem, we have z 4i z 4 6z 2 + dz = 4i 2πi [Resgz); z +) + Resgz); z )]) z = where gz) = z/z 4 6z 2 + ). We next find these residues: z + Resgz); z + ) = 4z 3 2z z + z=z+ = 4z+ 3 2z + = 4 z+ 2 3 = ) 3 = = 8 2. Since z + = z, we have their have the same squares. By examining the above calculation of the residue at z +, we find that the residue at z possesses the same value. Finally we now can evaluate the integral: 4i 2πi [Resgz); z + ) + Resgz); z )]) = 8π[2 Resgz); z + )] = 6π ) 8 2 = 2π = π