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+ for (a)0</zl</ (6) 12/> 1. -6) Find the two Laurent series in powers of z that represent sin --

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7:2 2 (-1) + 뚜 루튼 두 뜬 뚜루루 루 (주) (루시)E 니 Z w (나른 - (2) is not convergent. 1시리 ug mN 1시리 (4) -20 …… 2 - 2+1 - 루 - 도 발 1 - 0 장로Find Laurent sepies in powers of Z, of z2 sin 1 / 2 So, Sinz = IZKO Z - Z3 31 + 75 51 -1220+1 = n=o (2n+1)! sin 1/2 = 1 + + I

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