Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  nigelvr Dec 12 '12 at 2:52
    
Read here en.wikipedia.org/wiki/Laurent_series –  DonAntonio Dec 12 '12 at 2:54
add comment

1 Answer 1

up vote 1 down vote accepted

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

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

http://www.maths.manchester.ac.uk/~cwalkden/complex-analysis/complex_analysis_part6.pdf

share|improve this answer
    
+1 Argon - nice attachment! –  amWhy Jan 6 '13 at 1:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.