Question

(3) (5pt) Use multiplication of series to show that ) = $ +-3 2 +..., 0</st<1. BONUS (5pt): What is the Laurent expansion centered at 2 = i? In what region is this Laurent expansion valid?

Please solve the bonus!

Answer 1

series We begin by recalling the Malaurin representations, e?=1+ + 2+... t = 1+z+z²+ Z² + ... (17!<) ->(1) (IZIKI) and 1-Z So, (IZIC!) → (2), I 1+z2 h = 1 1 -z²+ Zt _ zot... 1-(-22) Muttaplying get, (1) and (2) term by term we = 1 +7 +32+ 2t - z - z3 -... + Z'+ ... = 1+ 2 - 1 2 2 2 ² 2 2 2 2 4 ... which is valid when Izlct. Then the desired lausent series is obtained as, - +1 -2-z+. (<1Z1<1) . Laurent expansion centered at z=i Suppose that a function of is analytic throughout an annular domain R, LIZ-i < R₂, centered at i and let e denote any positively oriented simple closed contour around zzi and lying in that domain Then at each point in the domain, BCZ) has the series representation, called Laurent expansion about is $(7) = Žancz-wynt See (Ro<!z -:1< Ps) 20

& (2) dz where (220, 1, 2, ...) 2π a Za-int save an o and bn 9(2) 60-012 (n=0,1,2,...) &(Z) az and nzo,1,2,... -nti ali J.