math185-hw2sol - MATH 185 COMPLEX ANALYSIS FALL 2007\/08 PROBLEM SET 2 SOLUTIONS Throughout the problem set i =-1 and whenever we write x yi it is

# math185-hw2sol - MATH 185 COMPLEX ANALYSIS FALL 2007/08...

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MATH 185: COMPLEX ANALYSIS FALL 2007/08 PROBLEM SET 2 SOLUTIONS Throughout the problem set, i = - 1; and whenever we write x + yi , it is implicit that x, y R . The real and imaginary parts of z C are denoted by Re( z ) and Im( z ) respectively, ie. if z = x + yi , then Re( z ) = x and Im( z ) = y . If Ω C , we let Ω R := { ( x, y ) R 2 | x + iy Ω } . 1. Suppose f ( z ) = n =0 a n z n has a radius of convergence R > 0. Let d be a fixed positive integer. Let g ( z ) = X n =0 n d a n z n and h ( z ) = X n =0 a n n ! z n . Find the radii of convergence of g ( z ) and h ( z ). Show that for any r with 0 < r < R , there is a constant M > 0 such that | h ( z ) | ≤ Me | z | /r . Solution. As we have discussed in the lecture, if ( x n ) n =1 is such that x n R and x n > 0 for all n N , then lim n →∞ n x n = lim n →∞ x n +1 /x n if the rhs exists. We may use this to easily determine the required superior limits. lim sup n →∞ | n d a n | 1 /n = lim n →∞ n n d lim sup n →∞ n p | a n | = lim n →∞ ( n + 1) d n d × 1 R = 1 R , lim sup n →∞ | a n /n ! | 1 /n = lim n →∞ n p 1 /n ! lim sup n →∞ n p | a n | = lim n →∞ n ! ( n + 1)! × 1 R = 0 . So the radius of convergence of g is R and the radius of convergence of h is . For the second part, let r (0 , R ). Then 1 /R < 1 /r . Since lim sup n →∞ n p | a n | = 1 R , there exists an N N such that | a n | ≤ 1 r n whenever n > N . Let M := max {| a n | r n | n = 0 , . . . , N } + 1 .

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