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.

  • $\begingroup$ This is not a "normal" power series but one called Laurent Series in complex analysis. Is this what you're up to? $\endgroup$
    – DonAntonio
    Dec 12, 2012 at 2:50
  • $\begingroup$ 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. $\endgroup$
    – nigel
    Dec 12, 2012 at 2:52
  • $\begingroup$ Read here en.wikipedia.org/wiki/Laurent_series $\endgroup$
    – DonAntonio
    Dec 12, 2012 at 2:54

1 Answer 1


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

$$f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n$$


  • 1
    $\begingroup$ Argon, the link no longer works. Can you post an updated link? $\endgroup$
    – Clyde
    Nov 15, 2015 at 17:02

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