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 a 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.