# How does the radius of convergence depend on the point about which the series is expanded?

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?

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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.

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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