Techniques for removing removable singularities (without resorting to series expansion)? Suppose $f: \mathbb{C} \supset U \to \mathbb{C}$ is a meromorphic function with a removable singularity at the point $z_0 \in U$.  Then $f$ can be extended to a holomorphic function over all of $U$.  However, the material I've encountered does not provide much in the way of practical techniques for explicitly computing the holomorphic extension.

The only technique I've encountered thus far is to simply write out the power series.  For instance, suppose we start with the function

$$ f(z) = \frac{e^z-1}{z}, $$

which has a removable singularity at $z=0$.  The corresponding power series is

$$ \frac{1}{z}\left( -1 + \sum_{n=0}^\infty \frac{z^n}{n!} \right) = \sum_{n=1}^\infty \frac{z^{n-1}}{n!}.$$

This series is certainly convergent (it is termwise smaller than the series for $e^z$), and is well-defined at $z=0$.  However, it leaves something to be desired: we started out with a nice, finite algebraic expression, and ended up with a nasty infinite sum.

One reason I call this sum "nasty" is that it does not lend itself well to numerical evaluation at an arbitrary point $z$.  Away from the origin, the original function $f(z)$ can be evaluated using a small number of standard numerical routines (exponentiation, division, etc.).  In contrast, to evaluate the sum to within machine precision, we likely have to evaluate a large number of terms (how many?), and we also have to be careful about things like catastrophic cancellation while accumulating the sum, especially near the origin.  What a mess!


  Question: Are there other techniques for explicitly constructing the holomorphic extension of a removable singularity, that do not
  result in series expansions?


(Note that in general I am interested in functions that are considerably more complicated than the example given above.)
The extension across the removable singularity simply coincides with the original function outside the singularitites (that's the very point!). At the removable singularity, the value of the extended function is just $\lim_{z\to a} f(z)$. In your particular case, we have that
$$
\lim_{z\to 0} \frac{e^z-1}{z} = 1,
$$
so the extended function is
$$
f(z) = \begin{cases} \frac{e^z-1}{z}, &amp; z \neq 0 \\ 1, &amp; z = 0 \end{cases}.
$$
The extension across the removable singularity simply coincides with the original function outside the singularitites (that's the very point!). At the removable singularity, the value of the extended function is just $\lim_{z\to a} f(z)$. In your particular case, we have that
$$
\lim_{z\to 0} \frac{e^z-1}{z} = 1,
$$
so the extended function is
$$
f(z) = \begin{cases} \frac{e^z-1}{z}, &amp; z \neq 0 \\ 1, &amp; z = 0 \end{cases}.
$$