Question

n=í (2n)! 2 ants (d) 0 2 ant} of these > #6 (18 pes) Find the Maclarin series of f1z)= 2²cos(22). (a) Ž L 22nt3 (-0, 3n+2 (5) não (2n)! (c) 2 nii (intl)! n! le) none #7 (15 pts) Find the Laurent series for 2444 in the region: 4<lz-zilco (5) Ź (2-zinte não lz-right2, (2i)" não 17-2i) 2, Cal 922–25" le) none of these hao la) (40) (c) Ř (Ztri)? no

Answer 1

6 Consider the function (z)= 73 C08 (22). As we know the Maclarin series for Cor 2: = g(z) is the loves E GIK z2k K=0 (2K)! Therefore the Macluarin series for cos (22) = g(22) is the series - 1)k zak (2K)! 2K)! GIJK (Z2) 2K k=0 K=0

Therefore for Macluaris series of f(z). we have f(x)= 23 COB (72) 73.g(72) 73 8 W (-)k z 4K kao (26) Gijk (2k)! z4K. 7 KED E GJk z 3+ 4k k=o (2k)!

។ Consider the function 1/2 2 44 in the region 4 L/Z-210 Which means. IL | 7 -231 4 → 4. La 17-251 Now consider ₃- (21)2 = (2 + 2;) (7-21) Z? +4 Now since (7-25) +2i +2; Z+2i

2- 2) +4) 1 "( (7-25) Г. + 41 і - 25 1 = Z-21) + 41 2 - 2ј Ая чое know 4 L1 1- 24) and thus 41 7-2 11

لما \1\Fox we have ... 13 ۔ 14 + 1-1 +ل رج 2 ZK k=0 Hence k .ارج أه 1 -2 F = 6 - 25 النه) کاری و k=0 (1-1). Thus. انه -1 (۱) (نة /21 انع + کره کرد =0 (2-2)۴

I Gigk (47) K (Z-2;)* (7-2) K=0 = Ž G-19K (43) * 7- 2i) k+1 k=0 Now. for th, we have Ž -2; - z² + 4 Z-23) (Z +23) Z-21) (Gisk (4;) k kao (7-2;) K** Ž + JU Gjk ('4;) * K-o ( 7-2;) ** (7-2) (ijk (4;)k (Z-23)+2 DO M8 K=0

Thus. for. 4. Iz-2; Lo we we have 8W 7² +4 h=0 Goh (4;)" 7- 23) 6+3 8 (-4;)" (7-7) 142


Thus option 1 is correct.