Disclaimer: I have virtually no complex analysis knowledge and I am seeking to get up to speed in this area of analysis. At the moment, I am surveying material on elementary complex functions and the related ideas pertaining to their singularities. I apologise if the questions I ask might be incredibly banal or obvious, but I really have very little knowledge in this are and am desperately seeking to better myself therein.
For $n\in\mathbb N$ and $\{a_n\in\mathbb F:n\in\mathbb N\}$ where $\mathbb F\in\{\mathbb R, \mathbb C\}$ consider the complex polynomial in $\frac{1}z$ given by $f:D(f)\subseteq\mathbb C\to\mathbb C$ where $f$ is defined
$$f(z):=\sum_{i=0}^na_nz^{-n}.$$ The domain of $f$, denoted $D(f)$, is clearly the set $\mathbb C/\{0\}$, since $f(z=0)$ is not well defined.
My question: is there any way to extend $f$ to all of $\mathbb C$? So as to have, for our above $f$, that $D(f)\equiv\mathbb C$? If so, how, precisely, does one go about "plugging in" the point $c=0$? (I have a feeling that this is to do with the notion of a singularity)
Unfortunately, it is impossible to extend $D(f)$ to include $0$ analytically in $\mathbb{C}$, as $\lim\limits_{z\to0}f(z)=\infty$ where $\infty$ really means that the limit diverges and does not exist. Thus plugging in any value for $f(0)$ leads to a jump discontinuity at $f(z=0)$.
In your comment, $g(z)=\frac{\sin(z)}{z}$ is able to be analytically extended as the limit $\lim\limits_{z\to0}f(z)=1$ does exist (in $\mathbb{C}$).