Radius of convergence of $\sqrt{z}$ about $4+3i$

Let \(f(z)\) be a branch of \(\sqrt{z}\) defined and holomorphic in a neighborhood of \(z_0 = 4 + 3i\) (for example, the principal branch).

1. The branch \(f\) is holomorphic on the disk \(\abs{z-z_0}<\abs{z_0}\), which does not contain the branch point \(0\), so the Taylor series of \(f\) about \(z_0\) has radius of convergence at least \[ |z_0| = |4 + 3i| = \sqrt{4^2 + 3^2} = 5. \]

2. Writing \(z = z_0 \left(1 + \frac{z - z_0}{z_0}\right)\), \[ \sqrt{z} = \sqrt{z_0} \left(1 + \frac{z - z_0}{z_0}\right)^{1/2} = \sqrt{z_0} \sum_{n=0}^\infty \binom{1/2}{n} \left(\frac{z - z_0}{z_0}\right)^n. \] Since \(1/2\notin\NN\), the ratio \(\abs{\binom{1/2}{n+1}/\binom{1/2}{n}}=\abs{1/2-n}/(n+1)\) tends to \(1\), so the binomial series \(\sum_{n=0}^\infty \binom{1/2}{n} w^n\) has radius of convergence exactly \(1\). With \(w = \frac{z - z_0}{z_0}\), the series converges for \(|z - z_0| < |z_0| = 5\) and diverges for \(|z-z_0|>5\).

Thus, the radius of convergence is \(R = 5\), the distance from \(z_0\) to the branch point \(0\).