Lecture 42:Uniqueness of Series RepresentationsDan SloughterFurman UniversityMathematics 39May 19, 200442.1 Uniqueness of Taylor seriesTheorem 42.1. Iff(z) =∞Xn=0an(z − z0)nfor all z in an open disk D = {z ∈ C : |z − z0| < R}, thenan=f(n)(z0)n!for n = 0, 1, 2, . . ..Proof. Let C be the circle |z − z0| = R1, where 0 < R1< R, and letgn(z) =12πi(z − z0)n+1,n = 0, 1, 2, . . .. ThenZCgn(z)f(z)dz =12πiZCf(z)(z − z0)n+1dz =f(n)(z0)n!.However, we also haveZCgn(z)f(z)dz =∞Xm=0am·12πiZC1(z − z0)n−m+1dz.1SinceZC1(z − z0)n−m+1dz =(0, if m 6= n,2πi, if m = n,it follows thatf(n)(z0)n!=ZCgn(z)f(z)dz = an.42.2 Uniqueness of Laurent seriesTheorem 42.2. Iff(z) =∞Xn=−∞cn(z − z0)nfor all z in an open annulus R0< |z − z0| < R1, R0≥ 0, thencn=12πiZCf(z)(z − z0)n+1dz,where C is any closed contour in the annulus with z0in its interior andn = 0, ±1, ±2, . . ..Proof. Similar to the previous proof, letgn(z) =12πi(z − z0)n+1,n = 0, ±1, ±2, . . .. ThenZCgn(z)f(z)dz =12πiZCf(z)(z − z0)n+1dz.However, we also haveZCgn(z)f(z)dz =∞Xm=−∞am·12πiZC1(z − z0)n−m+1dz.SinceZC1(z − z0)n−m+1dz =(0, if m 6= n,2πi, if m = n,2it follows thatcn=12πiZCf(z)(z −
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