Quotient of Complex Conjugates

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Theorem

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

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


Then:

$\overline {\paren {\dfrac {z_1} {z_2} } } = \dfrac {\paren {\overline {z_1} } } {\paren {\overline {z_2} } }$

for $z_2 \ne 0$.


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:

\(\ds \overline {\paren {\frac {z_1} {z_2} } }\) \(=\) \(\ds \overline {\paren {\frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} + i \frac {x_2 y_1 - x_1 y_2} { {x_2}^2 + {y_2}^2} } }\) Division of Complex Numbers
\(\ds \) \(=\) \(\ds \frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} - i \frac {x_2 y_1 - x_1 y_2} { {x_2}^2 + {y_2}^2}\) Definition of Complex Conjugate
\(\ds \) \(=\) \(\ds \frac {x_1 x_2 + \paren {-y_1} \paren {-y_2} } { {x_2}^2 + \paren {-y_2}^2} + i \frac {x_2 \paren {-y_1} - x_1 \paren {-y_2} } { {x_2}^2 + \paren {-y_2}^2}\)
\(\ds \) \(=\) \(\ds \frac {x_1 - i y_1} {x_2 - i y_2}\) Division of Complex Numbers
\(\ds \) \(=\) \(\ds \frac {\paren {\overline {z_1} } } {\paren {\overline {z_2} } }\) Definition of Complex Conjugate

$\blacksquare$


Sources