Radius of convergence is distance to the nearest singularity Can any one tell me please that whether Radius of Convergence of Power Series is distance to the nearest singularity from centre or distance to the nearest non-removable singularity to the centre? Or I want to know whether removable singularity matters or not?
Thanks
 A: Suppose $U$ is a non-empty open set in the complex plane, and $f$ is holomorphic in $U$. If $z_{0}$ is a point of $U$ and $\sum_{k=0}^{\infty}a_{k}(z - z_{0})^{k}$ is a power series representation of $f$ in some neighborhood of $z_{0}$, then the radius $R$ of this power series is the radius of the largest open disk $D(z_{0}, R)$ about $z_{0}$ into which $f$ (technically, the germ of $f$ at $z_{0}$) extends holomorphically.
Since $f$ can be extended holomorphically over a removable singularity, the presence of a removable singularity does not affect the radius $R$. For example, if $f(z) = \frac{\sin z}{z}$ for $z \neq 0$, then for every non-zero $z_{0}$, the power series expansion of $f$ about $z_{0}$ has infinite radius.
Caution: There is no guarantee we can expand $\frac{1}{z}$ and $\sin z$ separately and multiply series, since the series for $\frac{1}{z}$ has finite radius.
On a related note, if $\log$ denotes a holomorphic branch of logarithm in some region $U$, and if $z_{0}$ is a point of $U$ (necessarily non-zero), the series representation of $\log$ about $z_{0}$ has radius $|z_{0}|$, the distance to $0$. In this sense, points of a branch cut are not necessarily "singularities".
Caution: It need not happen that $f$ is equal to the power series in all of $D(z_{0}, R)$, only in the largest disk contained in $D(z_{0}, R) \cap U$.
