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A complex number $z$, in its nature, is nothing but an element $z \in \mathbb{R}^{2}$. Specifically, the imaginary unit $i$ is defined as $i = \left(0,1\right)$. Thus, our daily representation of a complex number $z = x + yi$ is abused in that, $x$ is a real number, and cannot be added to an element $yi \in \mathbb{R}^{2}$. In my understanding, $z = x + yi$ is actually the abbreviated form of $z = x\left(1,0\right) + yi = x\left(1,0\right) + y\left(0,1\right)$, where $\left(1,0\right)$ and $\left(0,1\right)$ are standard basis of $\mathbb{R}^{2}$. Thus, the $\mathrm{Re}\left(z\right)$ is the coefficient for $\left(1,0\right)$ while the imaginary is actually the coefficient for $\left(0,1\right)$. We should not see $i$ as something special. Is this understanding correct?

Ziqi Fan
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    It's kinda correct. The main difference is, $\mathbb C$ has a definition of multiplication that makes it into a field. $\mathbb R^2$ doesn't (only multiplication by a real). – Rushabh Mehta May 16 '21 at 00:48
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    What you describe is the isomorphism between $\mathbb R^2$ and $\mathbb C$ as real vector spaces. But there is more to $\mathbb C$ than just that, primarily the field structure. – dxiv May 16 '21 at 00:49
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    (Very) short answer: The important difference is 1. $\mathbb{C}$ is a field (which is algebraically closed) and 2. differentiability (see more in the answers to the post I linked to). Also your "main statement": "We should not see $i$ as something special." is opinion-based and not suited for math.se. – vitamin d May 16 '21 at 00:52
  • You should think of the complex numbers $\Bbb C$ as simply being a minimal field containing the real numbers and also an element $i$ with $i^2 =1$ ("minimal" meaning that it includes nothing else that is not required to meet those conditions). You should not consider it to have any other structure than what can be built from that description. $\Bbb R^2$ with the definitions you listed is a model of $\Bbb C$ - an explicit construction having all the properties of $\Bbb C$. However, there are other models for it. $\Bbb C$ itself is not any of the models, but a thing in and of itself. – Paul Sinclair May 16 '21 at 13:33
  • A couple examples of other models: $\Bbb R^2$ can be made a model in a different way: instead of $i = (0,1)$, you could instead make $i = (0,-1)$. It still works (which brings up a point: from the perspective of the real numbers, there is nothing that distinguishes $i$ from $-i$). – Paul Sinclair May 16 '21 at 13:40
  • Another model is to take the set $\Bbb R[x]$ of all polynomials in the variable $x$ with real coefficients. Then define two polynomials are equivalent if their difference is divisible by $x^2 + 1$. For each polynomial, the set of all polynomials equivalent to it is its "equivalence class". The set of all such equivalence classes forms a model of $\Bbb C$. The real numbers sit in it as the equivalence classes of constant polynomials. $i$ is the equivalence class of $P(x) = x$. But again, these are just models. $\Bbb C$ is not necessarily any model. – Paul Sinclair May 16 '21 at 13:46

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