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What differences -if any- are there between the complex numbers $\mathbb{C}$ and $\mathbb{R}^2$?

I am taking multi variable analysis now and I was wondering what possible changes there might be from the Analysis of two dimensions vs Complex Analysis.

A short summary of the differences would suffice.

naslundx
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StudentofEuler2718
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  • Complex analysis is concerned with complex-differentiable functions (mostly), these are much more regular than a real-differentiable function of two real variables usually is. – Daniel Fischer Mar 06 '14 at 15:35
  • Enormous. Hint: do you have a product in $\mathbb R^2$? – alex Mar 06 '14 at 15:35
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    The big difference is differentiability, which looks similar to differentiating functions $f: \mathbb{R}^2 \to \mathbb{R}^2$, but ends up becoming substantially different because unlike in $\mathbb{R}^2$, we have a well defined product operation which induces a multiplicative inverse for all non-zero elements. This changes the notion of differentiability -- actually, it induces a more restrictive definition -- which leads to classes of functions that have very interesting properties. See my answer here: http://math.stackexchange.com/a/444491/31475 – Emily Mar 06 '14 at 15:39

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Topologicaly they are the same, but algebraicaly not. Elements in $R^2$ and $C$ have different rules. Example in $R^2$ we have $(x_1,y_1)(x_2,y_2)=x_1x_2+y_1y_2$, but in $C$ we have $(x_1+y_1i)(x_2+y_2i)=x_1x_2-y_1y_2+(x_1y_2+x_2y_1)i$ or $(x_1,y_1)(x_2,y_2,)=(x_1x_2-y_1y_2,x_1y_2+x_2y_1)$.

Emo
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    In $\mathbb{R}^2$, there is no standard product, although the one you just defined is one of many possible products. – Emily Mar 06 '14 at 15:56
  • I disagree with you, because the 'natural' comparation of $C$ and $R^2$ can be made for the Euclidean (orthonormal) $R^2$. But still if you think that it can be for 'any' $R^2$, than we can put the general form of dot product: $xy=|x||y|cos\phi$ where $x,y \in R^2$ and $\phi$ is the angle between the coordinative vectors. So $xy=|x||y|cos\phi$ is still different from the product in $C$. – Emo Mar 06 '14 at 16:42
  • The dot product maps from $\mathbb{R}^2$ to $\mathbb{R}$. – Emily Mar 06 '14 at 16:47
  • Direct products from $R^2$ to $R^2$ in general come with a natural dot product: $(x_1,y_1)(x_2,y_2)=(x_1x_2,y_1y_2)$. – Emo Mar 06 '14 at 16:53
  • I hardly see that as a dot product, but it is, as you said, a direct product. Which is one possible natural product on $\mathbb{R}^2$, or $\mathbb{R}^n$ in general. But it has little to do with the "Euclidean" characteristics of $\mathbb{R}^2$, or those of $\mathbb{C}$. In fact, the direct product fails to satisfy the one Euclidean property we really care about: $| u\otimes v| = |u||v|$, viz., $$|u\otimes v| = \sqrt{(x_1 x_2)^2+(y_1y_2)^2} \neq \sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2} = |u||v|.$$ – Emily Mar 06 '14 at 17:03
  • You are right.
    But still that what you say doesn't deny my claim that $R^2$ and $C$ differ algebraically.     – Emo Mar 06 '14 at 17:11
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    It's true, they do differ algebraically. I don't dispute that. They differ because $\mathbb{C}$ has a standard product, where $\mathbb{R}^2$ does not. We could define $\mathbb{R}^2$'s product as $\mathbb{C}$'s product and have a complete isomorphism, though. – Emily Mar 06 '14 at 17:13
  • Bthw you've helped me to understand how we define products from $R^n$ to $R^m$. Thank you :) – Emo Mar 06 '14 at 17:17
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Think about this. In the definition of derivative of a function $f$ at a point $x_0$: $$\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h},$$ what doesn't make sense if $f:\mathbb R^2\rightarrow\mathbb R^2$ but does if $f:\mathbb C\rightarrow\mathbb C$?

alex
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