Suppose $f(z)= \frac{\sin(z)}{z^2 +64}.$ I want to argue that the radius of convergence of the power series expansion of $f$ at $z=6$ is $10.$ Since the function $f$ is holomorphic inside the open ball $\mathbb{B}(6, 10$) ( Open ball centered at 6 with radius 8), the function must be analytic on this region. Moreover, $10$ is the largest possible radius of such balls centered at $z=6.$ Is this argument correct? Thank you.
Radius of Convergence Exercise
analytic-functionscomplex-analysisconvergence-divergence
Related Solutions
More precisely, the radius of convergence is the radius of the largest open disk centred at the expansion point on which there is an analytic function that coincides with your function near the expansion point. If the series around $0$ has radius of convergence $R$, then the corresponding analytic function is analytic in the disk around $0$ of radius $R$, and therefore in the disk around $a$ of radius $R - |a|$.
However there is a loophole: who says that function is the same as the function you are expanding around $a$? You might have chosen a function that has a branch cut that comes between $0$ and $a$ (with branch point outside the big disk). If the branch cut had been chosen differently, you'd have a function analytic in the big disk. But the way you chose it, the function near $z=a$ is on a different branch, and this branch might have a singularity near $z=a$ that the other branch does not. For example, this can occur with a function of the form
$$ f(z) = \frac{1}{\sqrt{z-p} - q} $$ where $p$ is in the first quadrant and you choose the principal branch of the square root.
EDIT: For concreteness, let's take $$ f(z) = \frac{1}{\sqrt{z} + 1 - 10 i} $$ with the principal branch. It has a branch point at $0$. Note that the principal branch of $\sqrt{z}$ has real part $\ge 0$, so the denominator is never $0$. However, other branches may have a pole at $z = (1-10i)^2 = -99-20i$. The Taylor series around $z=-99-20i$ has radius $101$ (the distance to the branch point at $0$). But the Taylor series around $z=-99+20i$ has radius only $40$, because the analytic continuation in a disk around $-99+20i$ will run into a pole at $-99-20i$.
Best Answer
$f$ is analytic except for poles at $\pm8i$. Since $|6\pm8i|=10$, this shows that $f$ is analytic in $B(6,10)$ but not in $B(6,r)$ for $r>10$; hence the radius of convergence of the power series is $10$.