Condition on Conjugate from Real Product of Complex Numbers

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Theorem

Let $z_1, z_2 \in \C$ be complex numbers such that $z_1 z_2 \in \R_{\ne 0}$.


Then:

$\exists p \in \R: z_1 = p \overline {z_2}$

where $\overline {z_2}$ denotes the complex conjugate of $z_2$.


Proof

Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.

As $z_1 z_2$ is real:

$(1): \quad z_1 z_2 = x_1 x_2 - y_1 y_2$

and:

$(2): \quad x_1 y_2 + y_1 x_2 = 0$


So:

\(\ds \frac {\paren {z_1} } {\paren {\overline {z_2} } }\) \(=\) \(\ds \frac {x_1 + i y_1} {x_2 - i y_2}\) Definition of Complex Conjugate
\(\ds \) \(=\) \(\ds \frac {\paren {x_1 + i y_1} \paren {x_2 + i y_2} } {\paren {x_2 - i y_2} \paren {x_2 + i y_2} }\) multiplying top and bottom by $x_2 + i y_2$
\(\ds \) \(=\) \(\ds \frac {\paren {x_1 + i y_1} \paren {x_2 + i y_2} } { {x_2}^2 + {y_2}^2}\) Difference of Two Squares and $i^2 = -1$
\(\ds \) \(=\) \(\ds \frac {x_1 x_2 - y_1 y_2} { {x_2}^2 + {y_2}^2}\) from $(1)$


So $z_1 / \overline {z_2} = p$ where $p = \dfrac {x_1 x_2 - y_1 y_2} { {x_2}^2 + {y_2}^2}$, which is real.

$\blacksquare$


Sources

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