Complex Plane is Banach Space

Theorem

The complex plane, along with the complex modulus, forms a Banach space over $\C$.

Proof

The complex plane $\C$ is a vector space over $\C$.

That the norm axioms are satisfied is proven in Complex Modulus is Norm.

Then we have Complex Plane is Complete Metric Space.

Hence the result.

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces