Question

evaluate the following integrals. please show procedure.

4) 23. 01/22 dz Y a) r=1121=5), bydle-il-24 Sol: Ti r = {12-21 = 2 3 4 Sol: Ti 1 5) S dz 23(2-1) 4 r 6) J ze² z ²-1 dz 8=2 Izl=2) Sol: 2li cash (1) Y

Develop g(z)= 1/(z-1)(z-2) into a laurent series that is valid for the following anular domains.

Answer 1

Residue Theorem i Letr be a closed curve and f is analytic within and on v except finite no. of singularity within v tot, to ...etk. at then In f(2) dz = ani NS Res (40.); 2=t:)) at 2=20 if f(2) has a pole of order m, then Res (f(z) : 2= 20) = lim [(2-2)m feres 2+2 (me d dami so, if fly has a simple pole, of then, Res (f(2): 2=20) lim (2-0) f(2) 2720 s 3. e dz r 3 250 is the only singularity of function, dan so, in part @ & 6 answer will come same Res[ 2. et2 : 2=0] = co efficient of / ..:-[()+(397 - 2 + 2

Res[ 23. 6/22 : 220] 61- top therefore, sze : 22 v dz=ani [Res (zet - 25.2. 1 2 = (ni (in @ (5 both) 5 s teje dz = ? ra {12-21 = 3% v y OC can see easily, that only 2=1 point comes in v we will calculate residue at =1 only. 2=1 is a pole of order to de 271141 d23 1 . Res 1712-18 [12-14 i sin 23(2-1) 4 ) ding I 2+1 13 d23 line I 21 [1] 13 dele d d22 ling 1 21 B ding t 2 / 2 da = 3:12-13 4 1 2 ani [Res[ 23 (2434 Pri (10) - Froni

ze² zey da = ? r = {(2152 v clearly given function has simple peles at 2= 1 and 2-1, and both points of pole lies in given r. 80, we have to calculate Residue on both points. Res : 2 2ė (2+1) (24) 2-7). (2 + 1) (27) 2 2+1 وأما 271 and Res 2e² (2+1)(2-1) :2=1 [ (2+1). re zu 21 (2+1)(21) 1) ze² net lim 27-1 لا 2-1 de ze² Res de= 2e² 2=1 224 + Res re :2=-1 21 dai r + 9 ni (et) -sti dai cosh (1) zez dz=25i cosh(1) (21)