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 $$z_1 = x_1 + \imath y_1, z_2 = x_2 + \imath y_2$$.

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.