Complex Numbers form Vector Space over Reals

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

Let $$\mathbb{C}$$ be the set of complex numbers.

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

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

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

$$\forall x, y, \in \mathbb{C}, \forall \lambda, \mu \in \mathbb{R}$$:
 * 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 $$\lambda, \mu \in \mathbb{R}$$ it follows that $$\lambda, \mu \in \mathbb{C}$$ 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 $$\mathbb{C}$$.

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

VS 4 follows as $$1$$ is the unity of $$\mathbb{C}$$.