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How many infinities would be there for complex numbers? Like there are 2 infinities ($+\infty$ and $-\infty$) for the real numbers, is there a way to prove the number of infinities in the complex plane.

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up vote 4 down vote accepted

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.

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But that would mean that two functions going in completely different directions on the complex plane would finally meet at a point which is not making sense to me. Can you elaborate on the point? – sam_codes Feb 9 '14 at 14:18
Suppose we both sit at the North pole and we start walking along different meridians: we take different directions, but we will meet at the South pole. Actually from a topological point of view $\mathbb{C}\cup \{ \infty \}$ is just a sphere (have a look at things such as "Riemann sphere" or "stereographic projections"). – Lor Feb 11 '14 at 13:01
I think I understand it now. Thanks. – sam_codes Feb 11 '14 at 21:21

$-\infty$ 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.

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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 $\pm i\cdot\infty$.

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But using your method there would also be uncountable number of "zeroes" in the complex plane which we know is not true. – sam_codes Feb 9 '14 at 14:13

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