# Laurent series radius of convergence

Suppose one is given a series of the form $\sum\limits_{n=0}^\infty a_n (z - \alpha)^{-n}$ where $a_n,z,\alpha \in \mathbb{C}$ ($z$ is our indeterminate). How would one determine the radius of convergence of this series? Would the root (or ratio) test work? I imagine substituting $u = (z - \alpha)^{-1}$, looking at $\sum\limits_{n=0}^\infty a_n u^n$ and then applying the root or ratio test is the right thing to do, but I'm not sure.

• This is not a "normal" power series but one called Laurent Series in complex analysis. Is this what you're up to? – DonAntonio Dec 12 '12 at 2:50
• Yes, it is not a power series. I'll edit it to fix this. I wonder, however, if any of the normal tests for the radius of convergence apply. – nigel Dec 12 '12 at 2:52
• Read here en.wikipedia.org/wiki/Laurent_series – DonAntonio Dec 12 '12 at 2:54

Laurent series converge on the annulus $\left\{z\in \mathbb C| R_1 < |z − z_0| < R_2 \right\}$ where $0 \le R_1 < R_2 \le \infty$ when in the form
$$f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n$$