In complex analysis, we include infinity to complex set to extend it. Why we need to do this? And do we do the same for real and why so? And how complex plane and real plane related to compactification ( I don't know about it !)?


This may be a broad question to answer to, depending on what you need to know, but I'll try to get the important points. Let us start with $\mathbb{R}$: when you study functions there you're often asked to calculate limits at infinity, i.e. to $+\infty$ and $-\infty$. This makes sense because you only have two directions to check.
Let us try to do the same thing in $\mathbb{C}$: you now have infinite directions to check and, moreover, limits may be different when going to infinity on a particular curve, so it makes no sense to ask of "different" infinities. Though, we may be interested on how a function behaves when it goes to infinity in some fashion. A way to ask this is to consider when the function is very far from the origin, i.e. calculate the limit for $\|z \| \to \infty$. (*)

What does this mean geometrically? It means we are considering the complementary sets of the neighborhoods of the origin, when calculating the limit. This sounds pretty awkward because we are used to reason with neighborhoods when dealing with limits and not with their complementary sets. This can be overcome by extending the plane: $\hat{\mathbb{C}} = \mathbb{C} \cup \{ \infty \}$ and giving a topology to this new space. I won't give the details on how this new topology is defined, but if you're interested you should look for Alexandrov compactification or one-point compactification. But the main thing about this topology is that the complementary of compact sets (of course you need to add them the $\infty$) are neighborhoods of $\infty$. So, this topology translates geometrically the condition (*).

Addendum: a nice geometrical feature of this new topology is that it makes $\hat{\mathbb{C}}$ homeomorphic to $S^2$ (called Riemann sphere), so it is now a compact space.
Remember that you may do the same thing in $\mathbb{R}$ and it becomes homeomorphic to $S^1$. Though, usually $\mathbb{R}$ is not extended in that way but with a two-point compactification (using $-\infty, +\infty$), because we want to keep an order on the line on not use it like a circle. The first case corresponds to considering different limits at $+\infty$ and $-\infty$, the second correspond to considering the global limit for $|x| \to \infty$.

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  • $\begingroup$ Thanks for helping. I got it now. $\endgroup$ – Ka Sikh Apr 22 '17 at 12:13

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