Number of Infinities in complex numbers

Both $\mathbb{R}$ and $\mathbb{C}$ are non compact spaces, so it is natural to look for compactification and adding "points at infinity" is a way to compactify things. The way we decide to do it is in some sense arbitrary, a convention, which can of course be motivated or justified by what you want to do with the compactified space.

If you look at functions $f : \mathbb{R} \to \mathbb{R} $ you see that your function can diverge with higher and higher values or with lower or lower values, so it makes sense to add a point $+ \infty$ and a point $-\infty$ corresponding to the two different behaviours (different because you want to distinguish them). When you add these points your space becomes the same as a closed interval $[-\infty,+\infty]$ (i.e. is homeomorphic to it). You can do this essentially because $\mathbb{R}$ has a total order $\leq$ defined over itself, which allows you to distinguish an increasing function from a decreasing one.

If you forget about the order you can say that a function goes to infinity when its absolute value $|f|$ grows to infinity. Here you would need just one point at infinity, and the compactified space would look like a circle $S^1$.

When you look at complex functions $f : \mathbb{C} \to \mathbb{C} $, you have no order defined on the complex numbers (compatible with the algebraic structure), so you cannot talk about increasing/decreasing functions. The only thing that makes sense (keeping things easy) is to say that the absolute value of $f$ grows to infinity, so you add just one point at infinity, and you get a space which looks like a sphere $S^2$.

Of course there are other ways you could compactify things, for example you could add a point at infinity for each (real) direction you can take to go to infinity (i.e. leave every closed ball of $\mathbb{C}$), and this would give you the projective plane $\mathbb{R}\mathbb{P}^2$, but I don't know if this kind of completions is what you were looking for.

$-\infty$ and $\infty$ is the same point in the complex plane. It's actually not a part of the complex plane, but of the extended complex plane. It's more of a definition than anything else really, so "proving" the existence of the infinity point is kind of hard.

There are an uncountable number of “infinities” in the complex plane, of the form $\infty\cdot e^{i\alpha}$, where $\alpha\in[0,2\pi)$. See Euler's formula. For $\alpha=0$ and $\pi$ we have $\pm\infty$, and for $\alpha=\frac\pi2$ and $\frac{3\pi}2$ we have $\pm i\cdot\infty$.