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Unformatted text preview: Mathematics 185 Intro to Complex Analysis Fall 2009 M. Christ Solutions to Selecta from Problem Set 10 4.1.12 (This was a killer in terms of length.) Let f,g be analytic functions in an open punctured disk D = { z : 0 <  z z  < r } . Let them have Laurent series n = a n ( z z ) n and n = b n ( z z ) n , respectively. Prove that the infinite series c n = j = a j b n j converges for every n Z . Prove that n = c n ( z z ) n every z D , that its sum equals f ( z ) g ( z ), and that it is the Laurent series for fg in D . Solution, step 1. Consider first any two absolutely convergent numerical series n =0 A n and n =0 B n . I claim that if C n is defined to be the finite sum C n = n j =0 A j B n j for each n 0, then the series n =0 C n is absolutely convergent. For any n 0,  C n  n X j =0  A n   B n  . Therefore for any integer N 0, N X n =0  C n  X j,k N  A j   B k  . Indeed, the sum on the left is exactly the sum over all such pairs which satisfy the additional constraint j + k n . Therefore N X n =0  C n  X j,k N  A j   B k  = X j N  A j  X k N  B k  X j =0  A j  X k =0  B k  . Since these last two series converge, we have shown that the partial sums of the series n  C n  are uniformly bounded above by a certain finite quantity. Since n  C n  is a series with nonnegative terms, it converges. Thus n C n is absolutely convergent. Solution, step 2. I claim that X n =0 C n = X j =0 A j X k =0 B k . To prove this, consider 2 N X n =0 C n N X j =0 A j N X k =0 B k . The lefthand side is the sum of A j B k over all ordered pairs ( j,k ) of nonnegative integers satisfying j,k 2 N . The righthand is the sum of the same, over all ordered pairs satisfying the additional constraint j + k N . Therefore the difference is the sum of A j B k over certain 1 ordered pairs ( j,k ), all of which satisfy j + k > N . Therefore at least one of j,k must exceed N/ 2. Therefore  Big  2 N X n =0 C n N X j =0 A j N X k =0 B k X j>N/ 2 X k  A j   B k  + X j X k N/ 2  A j   B k  . AS N , each of the two terms on the righthand side tends to zero, since each is the sum of a tail of a convergent series. Therefore the lefthand side tends to zero. But the limit of the lefthand side equals n =0 C n j =0 A j k =0 B k , since the limit of a product is the product of the limits. Solution, step 3. Without loss of generality, we may assume that z = 0 to simplify notation. One nuisance is that the Laurent series for f,g are doubly infinite, whereas the above discussion applies only to singly infinite sums from n = 0 to n = + . To get around this define f M ( z ) = n = M a n z n and g M ( z ) = n = M b n z n for each integer M 0....
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 Fall '08
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