Writing Complex Numbers as a Vector in $\mathbb{R^2}$
Sophia Terry 
Generally speaking is it mathematically correct to write a complex number $z = x + iy \in \mathbb{C}$, as a vector in $\mathbb{R}^2$?
$$z=\begin{bmatrix} x\\ y \end{bmatrix} \quad \text{with } x,y \in \mathbb{R}$$
With row 1 being the $Re(z)$ and row 2 being $Im(z)$, i.e. $$Re: \mathbb R^2 \to \mathbb R, \begin{bmatrix} x\\ y \end{bmatrix}\mapsto x \quad \text{and} \quad Im: \mathbb R^2 \to \mathbb R, \begin{bmatrix} x\\ y \end{bmatrix}\mapsto y$$
$\endgroup$ 12 Answers
$\begingroup$Because the complex number space $\mathbb C$ is isomorphic to $\mathbb R^2$, the answer is yes. However, writing $(a+bi) \cdot (c+di)$ as
$$\begin{bmatrix} a\\ b \end{bmatrix} \cdot \begin{bmatrix} c\\ d \end{bmatrix}$$
does not make sense under normal vector multiplication - so you will need to define the multiplication operator in a different way.
Another way of doing this is to write complex numbers $a+bi$ as the matrix
$$\begin{bmatrix} a&-b\\ b&a \end{bmatrix}$$
which has the advantage of normal matrix-matrix multiplication giving the same result as normal complex multiplication.
$\endgroup$ 5 $\begingroup$Yes, the space $\mathbb C$ is isomorphic to the $\mathbb R^2$ with respect to the corresponding maps (addition, multiplication etc.).
It works like this: Consider the bijection
$$ \pi: \mathbb C \to \mathbb R^2, x+iy \mapsto \begin{bmatrix} x\\ y \end{bmatrix}$$ Now we define a fitting addition and multiplication on the $\mathbb R^2$. As addition we just take the entrywise addition, i.e.
$$\begin{bmatrix} x_1\\ y_1 \end{bmatrix} + \begin{bmatrix} x_2\\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2\\ y_1 + y_1 \end{bmatrix}$$ And we define a fitting multiplication, i.e.
$$\begin{bmatrix} x_1\\ y_1 \end{bmatrix} \cdot \begin{bmatrix} x_2\\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 x_2 - y_1y_2\\ x_1y_2 + x_2y_1 \end{bmatrix}$$ Now we can easily show that $\pi$ is a field isomorphism (left as exercise). Finally we can even define $Re$ and $Im$ like the following
$$Re: \mathbb R^2 \to \mathbb R, \begin{bmatrix} x\\ y \end{bmatrix}\mapsto x \quad \text{and} \quad Im: \mathbb R^2 \to \mathbb R, \begin{bmatrix} x\\ y \end{bmatrix}\mapsto y$$
This makes use of the fact, that $\mathbb C$ is a two dimensional $\mathbb R$-vector space (a base is for example $(1, i)$).
Hope that helps you :)
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