COMPLEX ANALYSIS: SOLUTIONS 5 1. Find the poles and residues of the following functions 1 1 π cot(πz) 1 ; ; ; (m; n 2 ) z4 + 5z2 + 6 (z2 − 1)2 z2 zm(1 − z)n Z>0 Solution: Throughout we use the following formula for calculating residues: If f(z) has a pole of order k at z = z0 then 1 dk−1 k res(f; z0) = k−1 (z − z0) f(z) : (k − 1)! dz z=z0 In particular, if f(z) has a simple pole at z0 then the residue is given by simply evaluating the non-polar part: (z −z0)f(z), at z = z0 (or by taking a limit if we have an indeterminate form). Let 1 1 1 f(z) := = = p p p p : z4 + 5z2 + 6 (z2 + 2)(z2 + 3) (z + i 2)(z − i 2)(z + i 3)(z − i 3) p p This has simple poles at z = ±i 2; ±i 3 with residues p p 1 1 1 res(f; i 2) = (z − i 2) p = p p = p ; z4 + 5z2 + 6 z=i 2 (z + i 2)(z2 + 3) z=i 2 2i 2 p p 1 1 res(f; −i 2) = (z + i 2) p = − p ; z4 + 5z2 + 6 z=−i 2 2i 2 p 1 res(f; i 3) = − p ; 2i 3 p 1 res(f; −i 3) = p : 2i 3 1 COMPLEX ANALYSIS: SOLUTIONS 5 2 For the second one let 1 1 f(z) = = : (z2 − 1)2 (z + 1)2(z − 1)2 This has double poles at ±1. From the formula we get d 1 res(f; 1) = = −1=4; dz (z + 1)2 z=1 d 1 res(f; −1) = = 1=4: dz (z − 1)2 z=−1 For the third let π cot(πz) f(z) = : z2 Now, cot(πz) has poles wherever sin(πz) = 0, so at z = n 2 Z. About these points we have (z − n)2 (z − n)3 sin(πz) = sin(πn) + π cos(πn)(z − n) − π2 sin(πn) − π3 cos(πn) + ··· 2! 3! (z − n)2 =(−1)nπ(z − n)1 − π2 + O((z − n)4) 3! and (z − n)2 (z − n)3 cos(πz) = cos(πn) − π sin(πn)(z − n) − π2 cos(πn) + π3 sin(πn) + ··· 2! 3! (z − n)2 =(−1)n1 − π2 + O((z − n)4): 2! Hence, for z close to n 2 Z, we have 1 − π2(z − n)2=2 + O((z − n)4) cot(πz) = π(z − n)1 − π2(z − n)2=6 + O((z − n)4) 1 − π2(z − n)2=2 + O((z − n)4) = 1 + π2(z − n)2=6 + O((z − n)4) π(z − n) 1 π = − (z − n) + O((z − n)3) π(z − n) 3 Therefore, f(z) = π cot(πz)=z2 has simple poles at z = n 6= 0 and a triple pole at z = 0. For the simple poles we have π cot(πz) 1 res(f; n) = (z − n) 2 = 2 : z z=n n COMPLEX ANALYSIS: SOLUTIONS 5 3 For the triple pole at at z = 0 we have 1 π2 1 f(z) = − + O(z) z3 3 z so the residue is −π2=3. Finally, the function 1 f(z) = zm(1 − z)n has a pole of order m at z = 0 and a pole of order n at z = 1. From the formula for residues we have m−1 1 d 1 res(f; 0) = m−1 n (m − 1)! dz (1 − z) z=0 n(n + 1) ··· (n + m − 2) = (m − 1)! (n + m − 2)! = (n − 1)!(m − 1)! and n−1 n 1 d (−1) res(f; 1) = n−1 m (n − 1)! dz z z=1 −m(m + 1) ··· (m + n − 2) = (n − 1)! (m + n − 2)! = − (m − 1)!(n − 1)! = − res(f; 0): 2. Use the substitution eiθ = z along with the residue theorem to show that Z 2π dθ 2π = p : 0 2 + cos θ 3 Solution: As suggested we let eiθ = z so that dθ = dz=(iz) and the integral becomes 1 Z dz 2 Z dz −1 = 2 : i jzj=1 z(2 + (z + z )=2) i jzj=1 z + 4z + 1 p 2 Now z + 4z + 1 has zeros of order 1 at z = z± = −2 ± 3 and so the integrand has simple poles at z+ and z−. Only z+ lies in the unit disk and therefore by the residue COMPLEX ANALYSIS: SOLUTIONS 5 4 theorem 2 Z dz 2 1 2 = × 2πi × res 2 ; z+ i jzj=1 z + 4z + 1 i z + 4z + 1 1 =4π(z − z+) z2 + 4z + 1 z=z+ 1 =4π(z − z+) (z − z )(z − z ) + − z=z+ 1 =4π z+ − z− 1 =4π p 2 3 2π =p : 3 3. Evaluate the following integrals via residues. Show all estimates. (i) Z 1 x2 4 2 dx 0 x + 5x + 6 (ii) Z 1 x sin x 2 2 dx; a real 0 x + a (iii) Z 1 log x 2 dx 0 1 + x Solution: (i) Since the integrand is an even function the integral in question is equal to I=2 where Z 1 x2 I = 4 2 dx: −∞ x + 5x + 6 p p As a function of a complex variable, the integrand has simple poles at ±i 2, ±i 3. We will be considering a semicircularp p contour in the upper half plane so we only need calculate the residues at z = i 2; i 3. A slight modification of the first calculation in question 1 gives p p z2 p (i 2)2 i 2 res ; i 2 = p = z4 + 5z2 + 6 2i 2 2 COMPLEX ANALYSIS: SOLUTIONS 5 5 and p p z2 p (i 3)2 i 3 res ; i 3 = − p = − : z4 + 5z2 + 6 2i 3 2 Now, Consider the semicircular contour ΓR, which starts at R, traces a semicircle in the upper half plane to −R and then travels back to R along the real axis. Then, on taking R large enough, by the residue theorem p p Z z2 X i 2 i 3 p p 4 2 dz = 2πi× residues inside ΓR = 2πi( − ) = π( 3− 2): ΓR z + 5z + 6 2 2 On the other hand Z z2 Z z2 lim dz = I + lim dz R!1 4 2 R!1 4 2 ΓR z + 5z + 6 γR z + 5z + 6 where γR is the semicircle in the upper half plane. But by the Estimation Lemma Z z2 z2 1 4 2 dz 6 πR max 4 2 ! 0 z2γR γR z + 5z + 6 z + 5z + 6 R p p as R ! 1. Hence, I = π( 3 − 2) and so Z 1 x2 π p p 4 2 dx = ( 3 − 2): 0 x + 5x + 6 2 The last estimate in the inequality for the integral over γR should be clear since the dominant term is the z4 term in the denominator, but for completeness: z2 π 1 πR max 4 2 = max −2 −4 z2γR z + 5z + 6 R z2γR 1 + 5z + 6z π 1 6 max −2 −4 R z2γR 1 − 5jzj − 6jzj π 1 = R 1 − 5R−2 − 6R−4 where we have used jz + wj > jzj − jwj in the second line. On taking R > 5, say, this is 6 2π=R and the constant 2π is absorbed into the symbol. (ii). We have Z 1 x sin x 1 Z 1 x sin x 1 Z 1 xeix 2 2 dx = 2 2 dx = = 2 2 dx : 0 x + a 2 −∞ x + a 2 −∞ x + a Denote this last integral by J. Again, we will consider J as the horizontal section of the contour ΓR from part (i). COMPLEX ANALYSIS: SOLUTIONS 5 6 In the upper half plane the integrand has a simple pole at z = ia with residue iz iz −a −a ze ze iae e res 2 2 ; ia = (z − ia) 2 2 = = : z + a z + a z=ia 2ia 2 Hence, by the residue theorem Z zeiz Z zeiz πie−a = lim dz = J + lim dz: R!1 2 2 R!1 2 2 ΓR z + a γR z + a Thus it remains to show that this last integral vanishes in the limit. This is similar to question 7 (ii) of Problems 3; a trivial estimate of the integrand is 1=R which is not enough for the Estimation Lemma. Instead we apply integration by parts which is probably the quickest way (see Problems 3). Integrating eiz and differentiating the rest gives Z iz iz −R Z 2 ze ze 1 1 2z iz 2 2 dz = 2 2 − 2 2 − 2 2 2 e dz: γR z + a i(z + a ) R i γR z + a (z + a ) The first term on the right is −2R cos R=i(R2 + a2) 1=R. For the integrals we use the Estimation Lemma to give Z iz iz e e 1 1 2 2 dz 6 πR max 2 2 6 πR 2 2 ; z2γR γR z + a z + a R − a R Z 2 iz 2 iz z e z e 3 1 1 2 2 2 dz 6 πR max 2 2 2 6 πR 2 2 2 ; z2γR γR (z + a ) (z + a ) (R − a ) R as R ! 1. Hence, J = πie−a and so Z 1 x sin x 1 π −a 2 2 dx = =(J) = e : 0 x + a 2 2 (iii) This is quite hard and, as I discovered recently, the solution is in Conway's book anyway (pg.