Question

1 Use a contour to evaluate So da 1+x4

Answer 1

Answer:- Given that Given integration is dac Itxu - let f (2) 17 z4+1 zut Put z+1=0 z z = yu (-1) z = 1/4 (cos Cos (Titan Titisin (Itant) = cos (2n+v) # tisin (20+1) for n=0,1,2,3 for noy Zocos os 1 y tisini ir + 1 3 for naiz, = -cos en tisingil :) + i sin stra for n=22 22= cos()+isina 2 for n=3 - 23 = 1 31- consider the count form as follows с N. żo

VSC (-top] Sf (2) d2 f (2) d2 + when Roo and f(2) is anaistic except By Jordan's lemma poles inside V, Then {f(z) dz=0 Rep { F(z);zo} = lim (z-zo) F (2) 22o (2-2) + 1424 423 lim zzo lim i zo 4.23 3 uzo Res{{{z); z;} lim 2->2, (2-2,) **(z) lim 2 -32, (2-2,)* - 1 1424 uz lim = 22,. U23 3 uz ${(2) d2 = SF(x)dx s f(x) dx = 21°{sum of residues} • 2ni {uza u23 + 2 To { 23 Gray Ja + 2

3 2 راما (5) 3 (+1)3 (2) 3 (1+1) 3 دا3 = تورهای + ) - X2 2 . t) 3( ز+ ۱-)