# Why the Euclidean topology in complex numbers?

This is a rather inconclusive question and I expect different answers:

The topology in $$mathbb {Q}$$ Y $$mathbb {R}$$ It is natural to be the order topology of these ordered fields.

Endowment $$mathbb {C}$$ with the topology of $$mathbb {R} ^ 2$$ is the key to all those good results of the classic complex analysis and the important fact, that a function $$mathbb {C} to mathbb {C}$$ It is differentiable if and only if it is a conformal map (preserving the angle locally). So the theory of holomorphic functions is more about geometry than $$mathbb {R} ^ 2$$ that in the properties of $$mathbb {C}$$ As a field, I think.

Now suppose, the existence of an algebraic closure of $$mathbb {Q}$$ It would have been discovered before the invention of $$mathbb {C}$$. It's hard to imagine that someone was like "let's endow $$mathbb {Q}[sqrt{-1}]$$ with the product metric of $$mathbb {Q} ^ 2$$ and embed its algebraic closure in its metric ending! "

Because there are so many other options: choose $$sqrt {2}$$ instead of $$sqrt {-1}$$; endow $$mathbb {Q}[sqrt{-1},sqrt{2}]$$ with the product topology of $$mathbb {Q} ^ 3$$; etc.

My question is: Why $$mathbb {C}$$ deserve the Euclidean topology of $$mathbb {R} ^ 2$$ and are there other options?