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Unformatted text preview: Mathematics 185 Intro to Complex Analysis Fall 2009 M. Christ Solutions to Selecta from Problem Set 9 3.2.13 Differentiate the power series (about 0) for (1- z )- 1 to obtain a power series expansion for (1- z )- 2 . What is the radius of convergence of the new series? Solution. (1- z )- 1 = n =0 z n , with radius of convergence 1. The derivative of (1- z )- 1 is (1- z )- 2 . Therefore in D (0 , 1), (1- z )- 2 = X n =1 nz n- 1 = X n =0 ( n + 1) z n . The radius of convergence of the new series is still 1. For instance, we can use the root test: ( n +1) 1 /n = e [ln( n +1)] /n , and [ln( n +1)] /n 0, so ( n +1) 1 /n e = 1. Therefore the radius of convergence is 1 = 1 / lim n ( n + 1) 1 /n . 3.2.14 Suppose that the power series n =0 a n z n has radius of convergence R . Let f ( z ) be the sum of this series, for | z | < R . Suppose that | z | < R . Let R be the radius of convergence of the Taylor series for f about z . Show that R- | z | R R + | z | . Solution. D ( z ,R- | z | ) D (0 ,R ). Indeed, if w is in this disk, then | w | = | ( w- z ) + z | | w- z | + | z | ( R- | z | ) + | z | = R. Therefore f is analytic in D ( z ,R- | z | ). Therefore its Taylor series expansion about z has radius of convergence R- | z | . Now the situation is symmetric, under reversal of the roles of the two centers z and 0. Since R R- | z | , 0 D ( z , R ). Therefore the radius of convergence R of the Taylor series about 0 satisfies R R- | z- | = R- | z | . Thus R R + | z | . 3.2.24 Compute the Taylor series of ( z ) = n =1 n- z about 2. Solution. Write n- z = e- z ln( n ) = e- 2 ln( n ) e- ( z- 2) ln( n ) = n- 2 X k =0 (- 1) k ln( n ) k k !...
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