Can we not we define every isolated singularity of a complex analytic function to be a removable singularity?

A removable singularity is a point where the holomorphic function is undefined, but it is possible to define the function at that point s.t. the function is regular in the neighborhood of that point.

So, most elemenary complex functions are analytic so say we had the function $f(z) = \frac{1}{(z^2 -1)}$. The isolated singularities are at $\pm 1$. Let's just look at the positive singularity. Can't we define $$h(z) = \begin{cases} f(z)& \text{if } z \neq 1\\ 2& \text{if } z = 1\end{cases}$$

I feel like when can near put any constant for the part in the piecewise function at the singularity point and then the function $h(z)$ will then be regular.

Can someone explain this?


The function won't be continuous, so it isn't a removable singularity:

$$ \lim_{z\to 1} h(z) \neq h(1)=2 $$

For a removable singularity the function must be bounded in the point's neighborhood, clearly not the case in this (and many other) cases.

An example for a removable singularity is $f(z)=\frac{\sin z}{z} $ at $z=0$. Naively, the function isn't defined at that point, but we can define $f(0)=1$ and the function will be continuous for every $z\in \mathbb{C}$ (and even analytic!)

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