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The Set of Complex numbers is a field as well as a nice topological space homeomorphic to $\mathbb{R}^2$. But why such a particular interest for this space? For instance what is more special about it than any other $\mathbb{R}^n$? I agree that each function on a complex domain will look like a single variable function and may have a nice definition for a derivative with respect to that single variable where as in $\mathbb{R}^3$ we cannot(?) have such a single variable to differentiate a function with.Particularly Holomorphic functions are as many times differentiable as we want. But is that all? I believe there is more to it but couldn't just get a picture of it. Any insights? Thanks

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    $\begingroup$ Any other $\mathbb R^n$ (for $n>2$) is not a field, to begin with. That's a pretty enormous difference in my book. Being the algebraic closure of $\mathbb R$ is no trifling matter, either. $\endgroup$ Sep 5, 2011 at 17:57
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    $\begingroup$ Complex analysis has a completely different feel to it than real analysis. $\endgroup$
    – KCd
    Sep 5, 2011 at 18:05
  • $\begingroup$ But a single complex variable is equivalent to two real variables, i.e., the homeomorphism; even diffeomorphism between the two is given $f: \mathbb C \rightarrow R^2 : z=x+iy \rightarrow (x,y)$ And, as Hardy mentioned, unlike $\mathbb R^2$, there is a product in $\mathbb C$, so that the complexes are an algebra over the reals .The definition of analitycity is also stricter than that of 'real-differentiability'; the first are analytic when they satisfy Cauchy-Riemann, but functions f:$\mathbb R^2 \rightarrow \mathbb R^2$ must only satisfy the limit definition of differentiability. $\endgroup$
    – gary
    Sep 5, 2011 at 19:30
  • $\begingroup$ It is one of few division algebras over the reals, the Hamiltonians, and Octonians making of the only other two. And they are not even commutative. $\endgroup$
    – Sven
    Sep 5, 2011 at 20:51
  • $\begingroup$ I just found out another question very relevant to this math.stackexchange.com/questions/5108/… $\endgroup$
    – Dinesh
    Sep 6, 2011 at 4:54

2 Answers 2

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I'll post this as an answer since I'm not allowed to comment.

As a response to your comment "I never knew multiplicative structure of C was responsible for holomorphicity", you have to use the field structure of the complex numbers to define the derivative via limits of quotients.

You might also wonder whether one can define such derivatives via quotiens in other $\mathbb{R}^n$ for n>2. However, in order to define a quotient one need an algebra structure such that any nonzero element has an inverse, ie, a division algebra structure. Due to classical theorems of Frobenius and Hurwitz this forces n=4 or 8 and corresponding algebra is not commutative (or even associative in the case n=8), which is undesirable if one wants do do calculus.

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  • $\begingroup$ thanks.. can you also give the reference for those $n=4,8$ theorems of Hurwitz and and Frobenius? $\endgroup$
    – Dinesh
    Sep 8, 2011 at 20:52
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    $\begingroup$ The Wikipedia articles are well written: Frobenius Theorem and Hurwitz Theorem $\endgroup$ Sep 12, 2011 at 20:19
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I think you got the main point: The definition of complex differentiable (holomorphic) functions uses the multiplicative structure of $\mathbb{C}$ in an essential way. It's not just a differentiable function from $\mathbb{R}^2 \to \mathbb{R}^2$. There are no higher-dimensional field extensions of $\mathbb{R}$, so we don't have an analog in higher dimensions.

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  • $\begingroup$ Well..I never knew multiplicative structure of $\mathbb{C}$ was responsible for holomorphicity. $\endgroup$
    – Dinesh
    Sep 5, 2011 at 18:36
  • $\begingroup$ You can define holomorphicity without the multiplicative structure but it isn't nearly as natural. $\endgroup$
    – Sven
    Sep 5, 2011 at 20:36

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