Product of Complex Conjugates

Theorem

Let $z_1, z_2 \in \C$ be complex numbers.

Let $\overline {z}$ be the complex conjugate of the complex number $z$.

Then:

$\overline {z_1 z_2} = \overline {z_1} \cdot \overline {z_2}$

Proof

Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \overline {z_1 z_2}\)	\(=\)	\(\displaystyle \overline {\left({x_1 x_2 - y_1 y_2}\right) + i \left({x_2 y_1 + x_1 y_2}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x_1 x_2 - y_1 y_2}\right) - i \left({x_2 y_1 + x_1 y_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x_1 x_2 - \left({-y_1}\right) \left({-y_2}\right)}\right) + i \left({x_2 \left({-y_1}\right) + x_1 \left({-y_2}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x_1 - i y_1}\right) \left({x_2 - i y_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \overline {z_1} \cdot \overline {z_2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$