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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $56 \ \text{(a)}$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): conjugate (of a complex number)