Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a given analytic germ $f(z)=\sum_{n=0}^{\infty} a_n (z-z_0)^n$, and a simple curve $\gamma: [0,1]\to \mathbb{C}$ such that $\gamma(0)=z_0$, suppose one may analytically continue $f(z)$ along $\gamma$. So for each point $\tilde{z} =\gamma(t)$ there is a corresponding power series $f(z;\tilde{z})=\sum_{n=0}^{\infty} \tilde{a}_n (z-\tilde{z})^n$ with its own radius of convergence $r(\tilde{z})$. My question is, what can be said about this function $r$? How regular is it? Is it continuous?

What about for more than one variable?

share|cite|improve this question
up vote 6 down vote accepted

The radius of convergence should be the distance to the nearest singular point. So it will be continuous, and it will be differentiable (in fact, smooth) except where its argument is equidistant from two or more singular points.

share|cite|improve this answer
Is there a similar result for $f$ a function of more than one complex variable? – AppliedSide Jan 20 '11 at 5:21
The radius of convergence can include singularities provided they are removable – Digital Gal Jan 20 '11 at 5:49

Your Answer


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.