Complex Multiplication Distributes over Addition

Theorem
The operation of multiplication on the set of real numbers $\C$ is distributive over the operation of addition.
 * $\forall z_1, z_2, z_3 \in \C:$
 * $z_1 \left({z_2 + z_3}\right) = z_1 z_2 + z_1 z_3$;
 * $\left({z_2 + z_3}\right) z_1 = z_2 z_1 + z_3 z_1$.

Proof
We use throughout the fact that the Real Numbers form Totally Ordered Field.

From the definition of complex numbers, we define the following:
 * $z_1 = x_1 + i y_1$
 * $z_2 = x_2 + i y_2$
 * $z_3 = x_3 + i y_3$

where $i = \sqrt {-1}$ and $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$.

Thus:

The result $\left({z_2 + z_3}\right) z_1 = z_2 z_1 + z_3 z_1$ follows directly from the above, and the fact that Complex Multiplication is Commutative.