# hw10 - Mathematics 185 – Intro to Complex Analysis Fall...

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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 left-hand side is the sum of A j B k over all ordered pairs ( j,k ) of nonnegative integers satisfying j,k ≤ 2 N . The right-hand 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 right-hand side tends to zero, since each is the sum of a tail of a convergent series. Therefore the left-hand side tends to zero. But the limit of the left-hand 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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## This note was uploaded on 01/19/2010 for the course MATH 113 taught by Professor Ogus during the Fall '08 term at Berkeley.

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hw10 - Mathematics 185 – Intro to Complex Analysis Fall...

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