In $C_{\infty}$ , does function $f(z) = \frac{1}{z}$ have removable singularity at $z=\infty$?

Question

In $C_{\infty}$ , does function $f(z) = \frac{1}{z}$ have removable singularity at $z=\infty$ ? Because $f(\frac{1}{w}) = \frac{1}{\frac{1}{w}}$ , $\frac{1}{w}$ isn't define at w = 0 and $lim_{w\to0}\frac{1}{\frac{1}{w}} = 0$.

Answer 1

Often, when we define arithmetic on $C_\infty$, we actually define division by $\infty$ (except for $\infty/\infty$), so $f$ is actually defined and continuous at $\infty$.

If you only define $f$ on the finite complex numbers, then you are correct, $f$ would have a removable singularity at $\infty$.