Question

(C)!!!!!

5. Find the Laurent series expansion of: 1 (a) f(x) = 1 about i, (b) f(x) = 22 + atz, convergent on {2< 121 < 4}, (c)* f(x) = 273-33+2, convergent on {{ < \z – 11 <1}.

Answer 1

I have done it for you in detail. Kindly go through.

3-32 5. (c) f(z) - (22-52+2) aze - 52 + 2 22 - 4 2 - 2+2 22 (2-2)-1(2-2) (22.-1) (2:2). so f(2)= 3(1-2) -3(2-1) (az-1) (2-2) (2(2-1)+1) (2-0-1) f(2) is analytic 12-11<!.. and Laurent series converges Ź </2-11<1. on Ź 2 B A Z-1 + [a 2-1) +1] [(2-1)-1] 262-1)+1 (2-1) - A {{Z-1)-1] HB [2(2-1)+] A (2-2) + B (22-1) Z-1 Z-1 -B-2A = and -1 2 B + A 28+ A=1 B + 2A!=ili and -48-2A-, -2 B + 2A = 1 -38 1-2/3 = 48 A = 1- 28 = B = 13 73 73 so - Z-1 2(2-1)+1 (2-1)-1 [2(2-1) +1][(2-1)-1]

* -3(2-1) [2 (2-1)+1] [[Z-1)-1] 2(2-1) + (2-1)-1 --) 2 Now 3 1 + 2 + 2 + 2 +.. So f(2) 1-E2(2-1) I-(2-) --[1-(-ate-))*4 [1- (2-1)] bave (1-2)' So f(2) - [1-(2-1)]'- [1-426-1377? {1+ + (2-1) +(2+(z=18+..}-{ 1+ (-2(2-1)) +62(2-1)} = {1+(2+)+(2+1)+(21)*}-{1-2 (2-1) + 4(21)8(2-1)} 0+3 (2-1) - 3(2-1)4+, 9(2+1) 8.- 15(20)44... Ź [1+tan) (3-1) So f(z) - 3(2-1) - 3(2-)+ 9 (2-1) 8-15 (2-1)4 .... 8 [1-6-2"] (2-1)" convergent on Ž </2-1/21 Y=0 m=0