Complex Plane is Metric Space

Theorem

Let $\C$ be the set of all complex numbers.

Let $d: \C \times \C \to \R$ be the function defined as:

$d \left({z_1, z_2}\right) = \left|{z_1 - z_2}\right|$

where $\left|{z}\right|$ is the modulus of $z$.

Then $d$ is a metric on $\C$ and so $\left({\C, d}\right)$ is a metric space.

Proof

Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.

From the definition of modulus:

$\left|{z_1 - z_2}\right| = \sqrt {\left({x_1 - x_2}\right)^2 + \left({y_1 - y_2}\right)^2}$

It is clear that this is the same as the euclidean metric, which is shown in Euclidean Metric is a Metric to be a metric.

Thus the complex plane is a 2-dimensional Euclidean space.

$\blacksquare$