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It is well known that $\mathbb{R}^2$ has a very famous field structure defiend by $(a,b)(c,d)=(ac-bd,ad+bc)$. And it also has a holomorphic structure, which makes $z\mapsto z$ differentiable but not $z\mapsto \bar{z}$. One of my friends asked me about the difference between $\mathbb{R}^2$ and $\mathbb{C}$, and I stated these two structures, but when I did it, I realized that I am not aware of the relation, or independence, of these two structures. So is one of these structures implies the other?

So, to elaborate my question more, it would be like this:

Suppose we have a holomorphic structure (of course with topology) on $\mathbb{R}^2$ which makes $(x,y)\mapsto(x,y)$ holomorphic but not $(x,y)\mapsto (x,-y)$. Now we want to find a field structure $\cdot:\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^2$ which is compatible with the holomorphic structure; $z \mapsto a\cdot z$ is complex-differentiable, and its derivative is a constant $a$. (edit) And of course the operation should be continuous with respect to the usual topology. Is it always isomophic to $\mathbb{C}$?

Here I stated the holomorphic structure first, but I am also curious about the other way: defining the field structure first and finding the 'compatible' holomorphic structure. And I am also curious if there is any way to see that one of these structures are actually implying the other 'directly,' which is the fact that I am not really sure of.

Thanks in advance!

(edit) I stated that the operation should be continuous with respect to the usual topology.

(edit) I changed "differentiability" condition into "holomorphicity."

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    You're confusing things a bit. The notion of "complex differentiation" does not correspond to a choice of a different differentiable structure on $\mathbb{R}^2$ in the sense of differential geometry. – 2017-02-07
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    Oh, or should I state it to be holomorphic structure? – 2017-02-07
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    Ah yea that would be better, should've probably understood that you meant that. – 2017-02-07
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    Thanks! i editted the question in case. – 2017-02-07
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    Just a small remark. Equip $\mathbb{R}^2$ with its usual topology. By the [uniformization theorem](https://en.wikipedia.org/wiki/Uniformization_theorem), there are exactly two distinct holomorphic structures on $\mathbb{R}^2$ (up to isomorphism): the standard one, and the one coming from the unit disk. This should simplify the question a fair bit. – 2017-02-07
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    As Jesse noted, an arbitrary choice of a complex structure on $R^2$ (equipped with the standard topology) does not allow for a compatible field structure. However, any continuous field structure on $R^2$ (no need for the holomorphic assumption) is isomorphic to the standard one, i.e. the field of complex numbers. – 2017-02-08

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