Question

Exercise 12: Residues and real integrals (a) [6+4 points) Compute the residues for all isolated singularities of the following functions (i) f(2)== (2-) tan(2), (i) 9(2):= z2 sin () (b) (4+6+5 points) Compute (using the Residue theorem) (1) cos(72) ( d, A3 := {z € C:<3), 243 := {Z EC: | = 3}, 34, (2-1)(2 + 2)2(2-4) : 43 € C:21 <3}, po 12 To (x2 + 4)2 da, 24 2 + 4 cosat. J 5 + 4 sin(t)

Answer 1

solution cosa foz) = (z-) tanz = (2-). Sina for point of singularity cos2 =0 = cos(ant + II ) 2=nitt I limi mut (2-5) bimeta = hence z= n + I is simple poleas of orders. Residues at a= hit + II. lim (2-(MA+II)) (2-II) Sinz { form. 3 2nT+777 = lim 1:12-1I). Sinz+(2-(nI+II)] 1. Sin2+(2-(n+1)(2-1), con - sinz Cosz 2 nI+I MIT Sin (npr TĮ) +0+0 - Sinin - ① DIT = – NT MER. g(2) = 6 2² sint for singular point z² sin =0 sinh Sint2=0 = sinnts tant 2-T to as nyo lim ze sind - - ith hence 220 ý limit point of the Iton hen i a=0 is isolated essential singularity. How Residue at 2=0, we expened sintz then find residue of f(a). coefficient of £ which is 2² sin = 29C 2 - 1 2 3 4 5 + 1 2 3 4 5 1 2 2 1/4 + ...] - 2- - Residue at 250 is coefficient of ² = -1

(b) solution i Costi z dz 20₂= {2EC: 121=3? Vans (2-1) (2+2)?(2-4) a = 1 and 2 is singular point for given contour. 2=1 is pole of order one and 2=-2 is pole of order two. Now Residue at z=1 lim (2-1) fcz) 2 - lim (23/073123°(2-43" a COST = (1+2) ? (174) = 127 L COSTI Residue at Z=-2 Residue at z=-2., lim lim – I you (2-1) (2725(2-4) z >-2 dz = lim 27-2 -sinti 2 7 (2-1)(2-4) - (22-5). Costi (2–11412-412 | = repole of orden mes 6-(-4-5) Cos 27 €3)2, (-6) ? to lim da 12-20) fz) = 104 27 2 2 = 36 By Cauchy's Residue theorem. &fizide=ani [Res-fers + Res frar] +2016+36] 5T 54 2 2 + + € zi in Cosno= 2+2, and sinna = " + the let fett = 24 Cost = 274.7 - 2 + 27 +2 etten - 5+4 sint - 5+47 stntot! - Sit-2 th ; al. for singular point 5t-ritzzi=0 (t + 1) (A +22) = 0 Here t= 1 1 is simple pole for given coantour oto 27. kesidue at t= i. lim. (4+1) fot = lim, A ti sti (k+di) =lim et tette - sit dt = 2Ti (Residue at t = 1) = 2Ti (2-3) = 41 g + 2T which is required. lim. (At il fut = lim (Ali2²+25+2 at ttri - 2 - i 2+ 4 cost sta sint