Question 6.23: Let f(z) = ln(1 + z), where we consider the branch that has ...
Let f(z) = ln(1 + z), where we consider the branch that has the zero value when z = 0. (a) Expand f(z) in a Taylor series about z = 0. (b) Determine the region of convergence for the series in (a). (c) Expand ln (1+z/1 – z) in a Taylor series about z = 0.
(a)
\begin{array}{rlrl}f(z) & =\ln (1+z), & f(0) & =0 \\f^{\prime}(z) & =\frac{1}{1+z}=(1+z)^{-1}, & f^{\prime}(0) & =1 \\f^{\prime \prime}(z) & =-(1+z)^{-2}, & f^{\prime \prime}(0) & =-1 \\f^{\prime \prime\prime}(z) & =(-1)(-2)(1+z)^{-3}, & f^{\prime \prime \prime}(0) & =2 !\\\vdots & & \vdots \\f^{(n+1)}(z) & =(-1)^n n !(1+z)^{-(n+1)}, & f^{(n+1)}(0) & =(-1)^n n !\end{array}Then
\begin{aligned}f(z)=\ln (1+z) & =f(0)+f^{\prime}(0) z+\frac{f^{\prime \prime}(0)}{2 !} z^2+\frac{f^{\prime \prime \prime}(0)}{3 !} z^3+\cdots \\& =z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\cdots\end{aligned}Another Method. If |z|<1,
\frac{1}{1+z}=1-z+z^2-z^3+\cdotsThen integrating from 0 to z yields
\ln (1+z)=z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\cdots(b) The nth term is u_n=(-1)^{n-1} z^n / n. Using the ratio test,
\lim _{n \rightarrow \infty}\left|\frac{u_{n+1}}{u_n}\right|=\lim _{n \rightarrow \infty}\left|\frac{n z}{n+1}\right|=|z|and the series converges for |z|<1. The series can be shown to converge for |z|=1 except for z=-1.
This result also follows from the fact that the series converges in a circle that extends to the nearest singularity (i.e., z=-1 ) of f(z).
(c) From the result in (a) we have, on replacing z by -z,
both series convergent for |z| < 1. By subtraction, we have
\ln \left(\frac{1+z}{1-z}\right)=2\left(z+\frac{z^3}{3}+\frac{z^5}{5}+\cdots\right)=\sum_{n=0}^{\infty} \frac{2 z^{2 n+1}}{2 n+1}which converges for |z| < 1. We can also show that this series converges for |z| = 1 except for z =±1.