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?