am i right in saying that the cartesian plane allows us to visualise elements of $\mathbb{R}^2$, and the complex plane elements of $\mathbb{C}$? My confusion was over whether the complex plane was a cartesian plane, technically, but i don't think it is.
Thanks
There is a natural identification between $\mathbb{C}$ and the plane $\mathbb{R}^2$:
$$a+bi \mapsto (a,b)$$
So technically, you can view $\mathbb{C}$ as the plane. This identification is made use of in many geometrical problems, as certain important geometrical transformations such as homothecy, rotation, reflection can be expressed as multiplication by complex numbers or complex conjugation.
The answer depend on how $\mathbb{C}$ is defined!
One way is as the quotient field $\mathbb{R}[X]/(X^2 + 1)$ of the ring of polynomials in one variable over $\mathbb{R}$ modulo the ideal generated by $X^2 + 1$, and then you may define the "identification" shown in the answer by A. Pongrácz, i.e., set up a bijection between the set of complex numbers and the set $\mathbb{R}^2$.
On the other hand, a simpler way is to define $\mathbb{C}$, as a set, to be the set $\mathbb{R}^2$ of ordered pairs of reals, and then that "identification" is in fact the identity map!
The distinction between $\mathbb{C}$ and $\mathbb{R}^2$ as algebraic systems comes only once you define the field operations on the first and the vector space operations on the second.
