expand the function f(z)=1/z^2 in a taylors series around z=1. What is the radius of convergence?
Attempted soln: Since 1/z^2 has a singularity at 0, isnt the radius of convergence just R=1-0 ?
You could do this:
$\displaystyle
f(x) = \frac{1}{z^2} = \frac{1}{1 + (z^2 - 1)}$
$\displaystyle \ \ \ \mid R \mid \leq 1 $
$\displaystyle \ \ \ R = (z^2 - 1) $
$\displaystyle \ \ \ \mid (z^2 - 1) \mid \ \ \leq 1
$