Understand definition of removeable singularity and extension of holomorphic function

Question

Hi I think I am lacking some knowledge to understand this.

Supppose $U\subset \mathbb{C}$ is open and $f$ is holomorphic on $U$ except at a point $p\in U$. Then $p$ is called an isolated singularity of $f$. We say $p$ is removeable if there exists a holomorphic function $g$ on $U$ that extends $f$. i,e. $f(z)=g(z)$ for all $z\in U$

Example, consdier $f(z)=\frac{e^z-1}{z}$ has a singularity at $z=0$ but it is removeable because $e^z-1=\sum^\infty _1 \frac{z^{n}}{n!}$, the power series $\sum \frac{z^{n-1}}{n!}$ extends $f$ on $\mathbb{C}$.

I dont understand what does the meaning of this extension over here, since $e^z-1=\sum^\infty _1 \frac{z^{n}}{n!}$ of course we have $f(z)$ equals that power series. Does this mean as long as you can write a function as a Power series, or we can manipulate the equation into a form where the function is well defined at the point of the singularity then their singularity is removeable? Or to put it very simple, you want the new "extend" in power series to have the value of $z^n$ to only have positive powers?

Thanks!

Answer 1

In OPs example we consider a function \begin{align*} &f:\mathbb{C}\setminus\{0\}\to\mathbb{C}\\ &f(z)=\frac{e^z-1}{z} \end{align*} which is holomorphic in $\mathbb{C}\setminus\{0\}$. At the point $z=0$ the function $f$ is undefined. It has an isolated singularity at that point.

Since \begin{align*} f(z)&=\frac{e^z-1}{z}=\frac{1}{z}\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)\\ &=1+\frac{z}{2}+\frac{z^2}{6}+\cdots\tag{1} \end{align*} we see the series (1) evaluated at $z=0$ is equal to $1$.

We can therefore define a function $g$: \begin{align*} &\color{blue}{g:\mathbb{C}\to\mathbb{C}}\\ &\color{blue}{g(z)=\begin{cases} f(z)&z\in\mathbb{C}\setminus\{0\}\\ 1&z=0 \end{cases}} \end{align*} This extension $g$ of $f$ is such that $f=g$ on $\mathbb{C}\setminus\{0\}$ and from (1) we also see that the definition $g(0)=1$ is a continuous in fact a holomorphic extension of $f$ with domain $\mathbb{C}$. The point $z=0$ is therefore a removable singularity of $f$.