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:

$\map d {z_1, z_2} = \size {z_1 - z_2}$

where $\size z$ is the modulus of $z$.

Then $d$ is a metric on $\C$ and so $\struct {\C, d}$ 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:

$\size {z_1 - z_2} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$

This is the euclidean metric on the real number plane.

This is shown in Euclidean Metric on Real Vector Space is Metric to be a metric.

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

$\blacksquare$


Sources

  • 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.4$
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