Find the first four terms in power-series expansion about the given point for the given function; find the largest disc in which the series is valid:
$$[Log(1-z)]^2 \ about \ z_0=0$$
Answer:
$$[Log(1-z)]' = -\frac{1}{1-z} = -\sum_{n=0}^{\infty}z^n, \text{ valid at } |z| <1$$
$$Log(1-z) =- \int\sum_{n=0}^{\infty}z^n dz =-\sum_{n=0}^{\infty}\int z^n dz = -\sum_{n=0}^{\infty}\frac{z^n+1}{n+1} =- \sum_{m=1}^{\infty}\frac{z^m}{m}$$
$$[Log(1-z)]^2 = [- \sum_{m=1}^{\infty}\frac{z^m}{m}]^2 = [\sum_{m=1}^{\infty}\frac{z^m}{m}]^2 \\=(z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+...)(z+\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+...) \\=z^2+z^3+\frac{11}{12}z^4+\frac{5}{6}z^5 +...$$
