Question

Solve:

Laurent series h(z) - Z O CIZ + 11 <3 (2+1)(2-2)

Answer 1

We have given hazl= ol1Z+K3 (2+1) CZ-2) Giren domain is os12+1R<3 Put 2 tl=t o 25t-1 2-22t-3 become o<lti<3 .; 06<+1133 Loki Z which become t-1 tct-3) have ha)= (2+1) (2-2) t-1 0x 1&< (2+1=t) (2+1)(2-2) Act 3 partial fraction ACT -3) ABE R Performing let t-1 trt-3) A t f t-3 +Ct-3) t-1 = Alt-3) + Bt At-3A ABE t-1 (A+B) E-3 Compare coefficerents of t & constant terms ATB El Ә) В -3A=4 H 3 mi a dlm 2 t- ... + = اے olts 3t 3 ct 30 +Ct-3) - 2 ochski lah 91-7-22

As ok Iles as = (5) 1-t 3 neo ㅈ t-1 ㅗ 3€ 2 9 ह ($) neo CZ+1) CZ-2) Łct-3) ¿ ť = 2 3+ 312 noo Z (2+ <+1)? 19 2 E (2+1) (2-2) 3(2+1) NEO 312 which is laurrent series expansion of hoz) in giren domain 0< Iz-t1133