Find the Taylor-series expansion of a square of a rational function of a complex variable

Question

I've been trying to find the Taylor-series expansion of the following function: $$ f(z)=\left ( \frac{1+z}{1-z} \right )^2 $$ az the origin : Z0 = 0. also I would like to find the region of convergence.

The problem I am having is to find the coefficients I must differentiate this function, and well I couldn't find the general formula for the derivative. The second path by evaluating the line integral also got me nowhere (using Cauchy's Integral Theorem).

This problem is from the book Mathematical Methods for Electrical Engineering by Thomas B.A. Senior (page 183).

Any help is appreciated.

Answer 1

Using algebraic manipulation and the geometric series you get $$ \left(\frac{1+z}{1-z}\right)^2 = \frac{(1+z)^2}{(1-z)^2}=\frac{1+2z+z^2}{1-z)^2}=\frac{1-2z+z^2 + 4z}{(1-z)^2} = 1+ \frac{4z}{(1-z)^2}$$ $$=1 + 4z\frac{d}{dz}\left(\frac{1}{1-z}\right) =1 + 4z\frac{d}{dz}\left(\sum_{k=0}^{\infty}z^{k}\right) =1 + 4z\sum_{k=1}^{\infty}kz^{k-1} = 1 + \sum_{k=1}^{\infty}4kz^k $$ This means that the Taylor coefficients $a_k$ for $f(z)=\sum_{k=0}^{\infty}a_kz^k$ are $$a_0=1, \quad a_k=4k \quad \text{for}\quad k>0$$ By the ratio test, the last series converges if $|z|<1.$

Answer 2

Note that $\left(\frac{1+ z}{1- z}\right)^2$ is defined everywhere except at z= 1. Since the "center" is z= 0, the radius of convergence is obviously 1.