$\mathbb{C}$ and $\mathbb{R}^{2}$

Question

If a complex number is a pair of real numbers, then why we need to introduce the new symbol $\mathbb{C}$ for complexes instead of using $\mathbb{R}^{2}$? What are the subtle differences involved?

Answer 1

Like you hint, the two spaces are isomorphic as (real) vector spaces. The difference is that when we write $\mathbb{C}$, we mean the underlying space together with the algebraic structure that makes $\mathbb{C}$ a field; beyond what is provided by underlying real vector space structure, this means we have a multiplication map $\mathbb{C} \times \mathbb{C} \to \mathbb{C}$ and inverse map $\mathbb{C} - \{0\} \to \mathbb{C} - \{0\}$ that satisfy various properties, e.g., commutativity and associativity of the multiplication map.

Remark 1 Note that $\mathbb{R}^2$ has another natural algebraic structure, namely the product algebra structure on $\mathbb{R} \times \mathbb{R}$ in which addition is the same as for $\mathbb{C}$ (namely, $(a, b) + (c, d) := (a + c, b + d)$) but multiplication is very different: $(a, b)(c, d) := (ac, bd)$. In particular, this algebra has zero divisors (e.g., $(1, 0)(0, 1) = (0, 0)$), so it is emphatically not a field.

Like $\mathbb{C}$, however, it has a natural conjugation map $\bar{\cdot}$ and a compatible nondegenerate $\mathbb{R}$-valued quadratic form, namely $N: x \mapsto x \bar{x}$ that respects the product structure, so that N(xy) = N(x)N(y)$; these properties make it a composition algebra, which is like an indefinite signature version of a normed division algebra.

Remark 2 One can define yet another natural algebraic structure on $\mathbb{R}^2$ by declaring that $i := (0, 1)$ satisfies $i^2 = +1$ (instead of $i^2 = - 1$), so that the product is given by $(a, b)(c, d) : = (ac + bd, ad + bc)$, and by analogy we call this algebra the split complex numbers $\widetilde{\mathbb{C}}$. Again this has zero divisors $(1, 1)(1, -1) = (0, 0)$. In fact this is isomorphic as an $\mathbb{R}$-algebra to $\mathbb{R} \times \mathbb{R}$ via the linear map $\mathbb{R} \times \mathbb{R} \to \widetilde{\mathbb{C}}$ characterized by $(1, 0) \mapsto (1, 1)$ and $(0, 1) \mapsto (1, -1)$.

Answer 2

$\mathbb C$ is a field in which you can both add and multiply the elements, i.e. for $z,w\in\mathbb C$, you have both $z+w$ and $zw$ in $\mathbb C$.

$\mathbb R^2$ is usually viewed as a vector space over the field $\mathbb R$, in which you have two operations: addition (which is identical to the addition in $\mathbb C$) and multiplication with scalars (like $\alpha\cdot (x,y)$) which is different.

Answer 3

I'm not a specialist bt my first thought is that because on $\mathbb{C}$ there is specific operation and behaviour with the introduction of $i^2=-1$ for example that we don't have in $\mathbb{R}^2$ And historicly, $\mathbb{C}$ was introduce to solve polynomial equation of degree 2.

Answer 4

The complex plane $\mathbb{C}$ differs from $\mathbb{R^2}$ in the fact of the imaginary element $i$. You can view $\mathbb{C}$ as the smallest field extension of $\mathbb{R}$ which contains both $\mathbb{R}$ and $i$, ie $\mathbb{C}=\mathbb{R}(i)$

Answer 5

If I understand your question, you are perfectly right that in a sense a complex number is equivalent to an ordered pair of real numbers. But when we look at arithmetic operations on complex numbers, we will see that for multiplication we need a very curious result: (x1,y1) (x2,y2) = (x1x2-y1y2,x1y2+x2y1). This is what the i notation conveniently does for us and why complex numbers are not simply ordered sets of reals.