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A holomorphic function $F$ on a larger open set $V$ whose restriction to $U$ is the original holomorphic function $f$. A removable singularity is a point across which such an extension exists. On a connected larger domain, a holomorphic extension is unique by the identity theorem, provided $U$ is nonempty.

Take the Laurent series of $f$ on $0<|z|<r$. For every integer $k\ge 1$, its coefficient of $z^{-k}$ satisfies, on a circle of radius $0<\rho <r$,The coefficient is independent of $\rho$. Letting $\rho \downarrow 0$ therefore gives $c_{-k}=0$ for every $k\ge 1$. The Laurent series is a power series with no negative powers and supplies a holomorphic extension at zero. Thus zero is a removable singularity. This is the Riemann removable singularity theorem, with boundedness used to eliminate the entire principal part.

The zeros of the hyperbolic sine are exactly $z=k\pi i$, $k\in \mathbb{Z}$, since $\sinh z=0$ is equivalent to $e^{2z}=1$. Each zero is simple because $\cosh (k\pi i)=(-1{)}^k\ne 0$. The quotient is therefore holomorphic precisely onAt zero, the Taylor expansion $\sinh z=z+O(z^3)$ gives $z/{\sinh }^{2}z=1/z+O(z)$, so zero is a simple pole. At $k\pi i\ne 0$, the denominator has a double zero and the numerator is nonzero, giving a pole of order two. Thus none of the omitted points admits a holomorphic extension.