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:

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

As $$\lambda, \mu \in \R$$ it follows that $$\lambda, \mu \in \C$$ and so $$(1)$$ and $$(2)$$ immediately follow from the fact that the complex numbers form a field.

So multiplication distributes over addition in $$\C$$.

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

$$(4)$$ follows as $$1 + 0 i$$ is the unity of $$\C$$.