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We know the two are of equal Cardinality $\mathbb c \times \mathbb c $, diffeomorphic as Real 2-manifolds, though I am not sure that Euclidean 2-Space is a Complex 1-manifold. The two are homeomorphic per the map $(x,y) \rightarrow x+iy $ I guess the Complexes are an algebra ( over the Reals) and the Euclidean Reals are not? I am also confused as to why/how z=x+iy is a single (Complex) variable: what "binds" x,y (both Real numbers) into the (single) variable $x+iy$? Thanks.

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  • $\begingroup$ idk what a,b are $\endgroup$ Jul 23 '19 at 17:39
  • $\begingroup$ What is your definition of $\mathbb C$? There are a few at en.wikipedia.org/wiki/Complex_number#Formal_construction $\endgroup$
    – lhf
    Jul 23 '19 at 17:47
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    $\begingroup$ The algebra point you made is exactly it. The complex numbers come equipped with a multiplication that give it a division algebra structure over $\mathbb{R}$. $\endgroup$
    – desiigner
    Jul 23 '19 at 17:47
  • $\begingroup$ @mathworker21: Sorry, just edited. $\endgroup$
    – MSIS
    Jul 23 '19 at 19:01
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As you say, they are indeed very similar objects! We're always thinking of ordered pairs of real numbers, $(x,y)$; the only difference is how we think about their structure. If you like, each of the definitions goes beyond the set of pairs, there is some baggage associated with it. When we speak of the complexes, the definition includes the field structure (the operations $+$, $-$, $\times$, $\div$ except by $0$). When we think of $\mathbb R^2$ we do it as a vector space, so there is a linear structure on it. The addition and subtraction operators are also present in the complexes, but now you have a scalar (dot) product, and often we also the matrix algebra ($2\times 2$ matrices) associated with linear maps.

Can you always turn complexes into $\mathbb R^2$ and vice versa? Of course! We just choose whichever setting is more convenient or natural. If you are thinking of Euclidean geometry or linear algebra, you might use vectors and matrices. If you are thinking of polynomial algebra, you might prefer to use the complexes. But there is nothing stopping you from expressing the complex-valued solutions to a quadratic equation as vectors in $\mathbb R^2$. It's just that many of us prefer the complex notation because that's how we were taught.

Hope this helps!

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You got it. $\mathbb C$ is an $\mathbb R$-algebra, while $\mathbb R^2$ is not (naturally speaking; c.f. Noah Schweber's comment). In fact you can define $\mathbb C$ as $\mathbb R^2$ with a fancy operation added onto it, namely $$(a,b)\star(c,d)=(ac-bd,ad+bc)$$

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    $\begingroup$ +1 - but there's an important subtlety here I want to point out for the OP. The set $\mathbb{R}^2$ is just that, the set of ordered pairs of real numbers. In that sense it's not an $\mathbb{R}$-algebra. $\mathbb{C}$ can be thought of as that same set equipped with some additional algebraic structure, per this answer. However, the statement "$\mathbb{R}^2$ is not [an $\mathbb{R}$-algebra]" is a bit dangerous: there is a canonical way to view it as an $\mathbb{R}$-algebra, namely as the direct product of $\mathbb{R}$ as an $\mathbb{R}$-algebra with itself. This algebra is "coordinatewise:" $\endgroup$ Jul 23 '19 at 17:54
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    $\begingroup$ Addition and multiplication are given by adding/multiplying like coordinates (meanwhile in $\mathbb{C}$, addition is coordinatewise but multiplication is the more complicated thing above). $\mathbb{R}^2$ and $\mathbb{C}$, as $\mathbb{R}$-algebras, are quite different: for example, in $\mathbb{R}^2$ we can multiply two nonzero things and get the zero element - think about $(0,1)$ and $(1,0)$ - while that can't happen in $\mathbb{C}$. $\endgroup$ Jul 23 '19 at 17:55
  • $\begingroup$ Great point, thanks $\endgroup$
    – Lance
    Jul 23 '19 at 17:56
  • $\begingroup$ @NoahSchweber: I am referring to Euclidean 2-space, not just R^2 as a set. My bad, I thought I had included that in the question, just edited title. $\endgroup$
    – MSIS
    Jul 23 '19 at 19:46
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    $\begingroup$ @MSIS Euclidean space is a space - that's it. There are then different algebraic structures which are "compatible," in the appropriate sense, with this space: the geometry does not uniquely determine the algebra. Put another way: the usual topologies on the $\mathbb{R}$-algebras $\mathbb{R}^2$ and $\mathbb{C}$ are homeomorphic. $\endgroup$ Jul 23 '19 at 21:18

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