2 n 2 2 1 2 2 2 n 1 1 n 1 1 2 n n 2 n 2 2 n 1 1 2 n 1 1 n 1 1 2 n n 2 n 2 2 n 1

# 2 n 2 2 1 2 2 2 n 1 1 n 1 1 2 n n 2 n 2 2 n 1 1 2 n 1

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(2 n - 2)! 2 · 1 · 2 · 2 · · · 2 · ( n - 1) = ( - 1) n - 1 1 2 n n ! (2 n - 2)! 2 n - 1 · 1 · 2 · · · ( n - 1) = = ( - 1) n - 1 1 2 n n ! (2 n - 2)! 2 n - 1 ( n - 1)! = ( - 1) n - 1 1 2 n n ! (2 n - 2)! 2 n - 1 ( n - 1)! 2 n (2 n - 1) 2 n (2 n - 1) = ( - 1) n - 1 1 2 n n ! (2 n )! 2 n n !(2 n - 1) ( - 1) n - 1 (2 n )!(2 n - 1) 2 2 n ( n !) 2 = ( - 1) n - 1 ( - 1) n - 1 2 2 n (2 n - 1) 2 n n . Thus (1 + z ) 1 2 = X n =0 ( - 1) n - 1 2 2 n (2 n - 1) 2 n n z n , | z | < 1 . 41. (a) and (b) There are several possible ways to derive the Taylor series expansion of f ( z ) = Log z about the point z 0 = - 1 + i . Here is one way. Let z 0 = - 1 + i , so | z 0 | = 2. The function Log z is analytic except on the negative real axis and 0. So it is guaranteed by
Section 4.3 Taylor Series 99 Theorem 1 to have a series expansion in the largest disk around z 0 that does not intersect the negative real axis. Such a disc, as you can easily verify, has radius Im z 0 = 1. However, as you will see shortly, the series that we obtain has a larger radius of convergence, namely | z 0 | = 2 (of course, this is not a contradiction to Theorem 1). Consider the function Log z in B 1 ( z 0 ), where it is analytic. (The disk of radius 1, centered at z 0 is contained in the upper half-plane.) For z B 1 ( z 0 ), we have d dz Log z = 1 z . Instead of computing the Taylor series of Log z directly, we will first compute the Taylor series of 1 z , and then integrate term-by-term within the radius of convergence of the series (Theorem 3, Sec. 4.3). Getting ready to apply the geometric series result, we write 1 z = 1 z 0 - ( z 0 - z ) = 1 z 0 ( 1 - z 0 - z z 0 ) ( z 0 6 = 0) = 1 z 0 · 1 1 - z 0 - z z 0 = 1 z 0 X n =0 z 0 - z z 0 n = 1 z 0 X n =0 ( - 1) n ( z - z 0 ) n z n 0 , where the series expansion holds for z 0 - z z 0 < 1 ⇔ | z 0 - z | < | z 0 | . Thus the series representation holds in a disk of radius | z 0 | = 2, around z 0 . Within this disk, we can integrate the series term-by-term and get Z z z 0 1 ζ = 1 z 0 X n =0 ( - 1) n z n 0 Z z z 0 ( ζ - z 0 ) n = 1 z 0 X n =0 ( - 1) n ( n + 1) z n 0 ( z - z 0 ) n +1 . Reindexing the series by changing n + 1 to n , we obtain Z z z 0 1 ζ = X n =1 ( - 1) n +1 nz n 0 ( z - z 0 ) n | z - z 0 | < | z 0 | = 2 . Now we have to decide what to write on the left side. The function Log z is an antiderivative of 1 z in the disk of radius 1, centered at z 0 . (Remember that Log z is not analytic on the negative real axis, so we cannot take a larger disk.) So, for | z - z 0 | < 1, we have Z z z 0 1 ζ = Log ζ z z 0 = Log z - Log z 0 . Thus, for | z - z 0 | < 1, we have Log z = Log z 0 + X n =1 ( - 1) n +1 nz n 0 ( z - z 0 ) n , even though the series on the right converges in the larger disk | z - z 0 | < 2.
100 Chapter 4 Series of Analytic Functions and Singularities Solutions to Exercises 4.4 1. We have that 1 1 + z = 1 z 1 1 + 1 z = 1 z X n =0 - 1 z n = X n =0 ( - 1) n z n +1 , and this converges provided - 1 z < 1, that is, when 1 < | z | . 5. Since 1 z 2 < 1, we can use a geometric series in 1 z 2 as follows. We have 1 1 + z 2 = 1 z 2 1 1 - - 1 z 2 = 1 z 2 X n =0 - 1 z 2 n = 1 z 2 X n =0 ( - 1) n z 2 n = X n =0 ( - 1) n z 2( n +1) = X n =1 ( - 1) n - 1 z 2 n , where in the last series we shifted the index of summation by 1. Note that ( - 1) n - 1 = ( - 1) n +1 , and so the two series that we derived are the same.
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