Complex Multiplication is Commutative

From ProofWiki
Jump to: navigation, search

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:

$z = \left({x_1, y_1}\right)$
$w = \left({x_2, y_2}\right)$

where $x_1, x_2, y_1, y_2 \in \R$.


Then:

\(\displaystyle z_1 z_2\) \(=\) \(\displaystyle \left({x_1, y_1}\right) \left({x_2, y_2}\right)\) $\quad$ Definition 2 of Complex Number $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1}\right)\) $\quad$ Definition of Complex Multiplication $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({x_2 x_1 - y_2 y_1, x_1 y_2 + x_2 y_1}\right)\) $\quad$ Real Multiplication is Commutative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({x_2 x_1 - y_2 y_1, x_2 y_1 + x_1 y_2}\right)\) $\quad$ Real Addition is Commutative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({x_2, y_2}\right) \left({x_1, y_1}\right)\) $\quad$ Definition of Complex Multiplication $\quad$
\(\displaystyle \) \(=\) \(\displaystyle z_2 z_1\) $\quad$ Definition 2 of Complex Number $\quad$

$\blacksquare$


Sources