Understanding Proof on Analytic Continuation .

99 Views Asked by At

In Book Complex Analysis by Stain and Shakarchi .
I read following proof .
enter image description here

enter image description here

I understand that for $f\neq 0$ then there exist some non zero m such that $a_m\neq 0$ but I do not understand why f(z) can be written as above highlighted form.
Any Help will be appreciated

1

There are 1 best solutions below

0
On BEST ANSWER

Becaue\begin{align}f(z)&=a_m(z-z_0)^m+a_{m+1}(z-z_0)^{m+1}+\cdots\\&=a_m(z-z_0)^m\left(1+\frac{a_{m+1}}{a_m}(z-z_0)+\frac{a_{m+2}}{a_m}(z-z_0)^2+\cdots\right).\end{align}So, let$$g(z)=\frac{a_{m+1}}{a_m}(z-z_0)+\frac{a_{m+2}}{a_m}(z-z_0)^2+\cdots$$