Complex Multiplication is Commutative

Theorem

The operation of multiplication on the set of complex numbers $\C$ is commutative:

$\forall z_1, z_2 \in \C: z_1 z_2 = z_2 z_1$

Proof

From the definition of complex numbers, we define the following:

$z_1 = x_1 + i y_1$

$z_2 = x_2 + i y_2$

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

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle z_1 z_2\)	\(=\)	\(\displaystyle \left({x_1 + i y_1}\right) \left({x_2 + i y_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x_1 x_2 - y_1 y_2}\right) + i \left({x_1 y_2 + x_2 y_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of complex multiplication
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x_2 x_1 - y_2 y_1}\right) + i \left({x_2 y_1 + x_1 y_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Real Addition is Commutative and Real Multiplication is Commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x_2 + i y_2}\right) \left({x_1 + i y_1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of complex multiplication
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle z_2 z_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$