Let's take an analytical function $f(x) = \sum_{-\infty}^{\infty}c_n(z-a)^n$. It may be a Laurent series as well. Let's also hypothesize that it converges in the Disc $D(a,r,R)$. If I attempt to calculate the power series with a new center what will happen to this disc? Can you give also an example? I had an intuitive idea that if I change the center to $b$ then $R=\operatorname{dist}(b,O^*)$ where $O^*=\{z: f(z) \text{ is not holomorphic at $z$}\}$. Am I wrong? Is there a proof you can redirect me to?

NOTE: $\operatorname{dist}(z,A) = \inf \operatorname{dist}(z,a)$ where $z \in A$

  • $\begingroup$ Is this your question? math.stackexchange.com/q/1732356/451957 $\endgroup$ – Epiousios Sep 23 '17 at 18:33
  • $\begingroup$ No sorry, I have read all related articles I could find here but they do not answer my question $\endgroup$ – Pookaros Sep 23 '17 at 19:31

Suppose $a$ is a singularity of $f$, so $f$ converges on some annulus

$$D(a,r,R) = \{ z : r < |z| < R \}$$

If $z_0 \in D(a,r,R)$, then $f$ has a power series expansion about $c$ as

$$f(z) = \sum\limits_{n=0}^{\infty} c_n(z-z_0)^n$$

where $c_n = \frac{f^{(n)}(z_0)}{n!}$. As I understand, you want to know what can be said about the radius of convergence of this power series.

The only thing that can be said for sure is that this radius is greater than or equal to the minimum of thes distance from $z_0$ to the inner circle and the outer circle.

The behavior of $f$ near the inner circle of the annulus can be quite bad. Or it could extend within the inner circle but have other singularities in there besides at $a$. If $f$ extends to an analytic function on all of $D(a,r) - \{a\}$, with a pole or essential singularity at $a$, then the radius of convergence of the power series can be no greater than $|z_0 - a|$. It also might be shortened by the distance from $z_0$ to the edge of the outer circle.


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