Find the radius of convergence of the Taylor series of the principal branch of \(\sqrt z\) about \(z_0=4+3i\). Repeat with \(z_1=-4+3i\).
Solution
For a center \(c\neq0\), the disk \(\abs{z-c}<\abs c\) does not contain \(0\), so it carries a holomorphic branch of \(\sqrt z\) agreeing with the principal branch near \(c\); the Taylor series of the principal branch at \(c\) converges on this disk. It converges on no larger disk: the binomial expansion \[ \sqrt z=\sqrt c\sum_{n\ge0}\binom{1/2}{n}\qty{z-c\over c}^n \] has radius of convergence exactly \(\abs c\), since \(\binom{1/2}{n}\) has ratio \(\abs{1/2-n}/(n+1)\to1\).
For \(z_0=4+3i\), \[ R_0=|4+3i|=\boxed{5}. \]
For \(z_1=-4+3i\), \[ R_1=|-4+3i|=\boxed{5}. \] The branch cut \(\RR_{\le0}\) meets this disk: the sum of the series equals the principal branch only on the part of the disk above the cut, including the disk \(\abs{z-z_1}<3\), and equals its analytic continuation across the cut below it.