Complex Numbers form Vector Space over Reals

Theorem
Let $$\R$$ be the set of real numbers.

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

Then the $\R$-module $$\C$$ is a vector space.

Proof
First note that $$\R$$, being a field, is also a division ring.

Thus we only need to show that $\R$-module $$\C$$ is a unitary module, by demonstrating the module properties:

$$\forall x, y, \in \C, \forall \lambda, \mu \in \mathbb{R}$$:
 * VS 1: $$\lambda \left({x + y}\right) = \left({\lambda x}\right) + \left({\lambda y}\right)$$;
 * VS 2: $$\left({\lambda + \mu}\right) x = \left({\lambda x}\right) + \left({\mu x}\right)$$;
 * VS 3: $$\left({\lambda \mu}\right) x = \lambda \left({\mu x}\right)$$.
 * VS 4: $$1 x = x$$.

As $$\lambda, \mu \in \R$$ it follows that $$\lambda, \mu \in \C$$ and so VS 1: and VS 2: immediately follow from the fact that the Complex Numbers form a Field, and so multiplication distributes over addition in $$\C$$.

VS 3 follows from the fact that multiplication is associative on $$\C$$, again because $$\C$$ is a field.

VS 4 follows as $$1 + 0 \imath$$ is the unity of $$\C$$.