Let $f(z)=\frac{(z^4-1)^2\sin^2(z)}{\cos(2\pi z)-1}$ and let $U$ be the maximal domain of $f$. Let $V\subseteq \mathbb{C}$ be the maximal set on which a holomorphic $g$ can be defined that agrees with $f$ on $U$. Determine the radius of convergence of the power series of $g$ at each of the following points $$ z=i\qquad z=1+i\qquad z=2+i\qquad z=3+i $$
I found that $U=\mathbb{C}\setminus\mathbb{Z}$ and the singularties in $z=-1,0,1$ are removable singularities of $f$, hence $V=U\cup\{-1,0,1\}$. Then I thought that radius of convergence of $g$ around $z=i$ is the distance to $2$, hence is $\sqrt{5}$. Around $z=1+i$ the radius of convergence is the distance to $2$, hence is $\sqrt{2}$, and around $z=2+i$ and $z=3+i$ the radius of convergence is 1. Is this method correct?