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(2n-2)!2·1·2·2· · ·2·(n-1)= (-1)n-112nn!(2n-2)!2n-1·1·2· · ·(n-1)== (-1)n-112nn!(2n-2)!2n-1(n-1)!= (-1)n-112nn!(2n-2)!2n-1(n-1)!2n(2n-1)2n(2n-1)=(-1)n-112nn!(2n)!2nn!(2n-1)(-1)n-1(2n)!(2n-1)22n(n!)2= (-1)n-1(-1)n-122n(2n-1)2nn.Thus(1 +z)12=∞Xn=0(-1)n-122n(2n-1)2nnzn,|z|<1.41.(a) and (b)There are several possible ways to derive the Taylor series expansion off(z) = Logzabout the pointz0=-1 +i. Here is one way. Letz0=-1 +i, so|z0|=√2.The function Logzis analytic except on the negative real axis and 0. So it is guaranteed by
Section 4.3Taylor Series99Theorem 1 to have a series expansion in the largest disk aroundz0that does not intersectthe negative real axis. Such a disc, as you can easily verify, has radius Imz0= 1. However,as you will see shortly, the series that we obtain has a larger radius of convergence, namely|z0|=√2 (of course, this is not a contradiction to Theorem 1).Consider the function LogzinB1(z0), where it is analytic.(The disk of radius 1,centered atz0is contained in the upper half-plane.) Forz∈B1(z0), we haveddzLogz=1z.Instead of computing the Taylor series of Logzdirectly, we will first compute the Taylorseries of1z, 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 write1z=1z0-(z0-z)=1z0(1-z0-zz0)(z06= 0)=1z0·11-z0-zz0=1z0∞Xn=0z0-zz0n=1z0∞Xn=0(-1)n(z-z0)nzn0,where the series expansion holds forz0-zz0<1⇔ |z0-z|<|z0|.Thus the series representation holds in a disk of radius|z0|=√2, aroundz0. Within thisdisk, we can integrate the series term-by-term and getZzz01ζdζ=1z0∞Xn=0(-1)nzn0Zzz0(ζ-z0)ndζ=1z0∞Xn=0(-1)n(n+ 1)zn0(z-z0)n+1.Reindexing the series by changingn+ 1 ton, we obtainZzz01ζdζ=∞Xn=1(-1)n+1nzn0(z-z0)n|z-z0|<|z0|=√2.Now we have to decide what to write on the left side. The function Logzis an antiderivativeof1zin the disk of radius 1, centered atz0. (Remember that Logzis not analytic on thenegative real axis, so we cannot take a larger disk.) So, for|z-z0|<1, we haveZzz01ζdζ= Logζzz0= Logz-Logz0.Thus, for|z-z0|<1, we haveLogz= Logz0+∞Xn=1(-1)n+1nzn0(z-z0)n,even though the series on the right converges in the larger disk|z-z0|<√2.
100Chapter 4Series of Analytic Functions and SingularitiesSolutions to Exercises 4.41.We have that11 +z=1z11 +1z=1z∞Xn=0-1zn=∞Xn=0(-1)nzn+1,and this converges provided-1z<1, that is, when 1<|z|.5.Since1z2<1, we can use a geometric series in1z2as follows. We have11 +z2=1z211--1z2=1z2∞Xn=0-1z2n=1z2∞Xn=0(-1)nz2n=∞Xn=0(-1)nz2(n+1)=∞Xn=1(-1)n-1z2n,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.