Product of Complex Number with Conjugate by Dot and Cross Product
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Theorem
Let $z_1$ and $z_2$ be complex numbers.
Then:
- $\overline {z_1} z_2 = \paren {z_1 \circ z_2} + i \paren {z_1 \times z_2}$
where:
- $\overline {z_1}$ denotes the complex conjugate of $z_1$
- $z_1 \circ z_2$ denotes the complex dot product of $z_1$ with $z_2$
- $z_1 \times z_2$ denotes the complex cross product of $z_1$ with $z_2$.
Proof
\(\ds \) | \(\) | \(\ds \paren {z_1 \circ z_2} + i \paren {z_1 \times z_2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {z_1 \circ z_2} + i \frac {\overline {z_1} z_2 - z_1 \overline {z_2} } {2 i}\) | Definition 4 of Vector Cross Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\overline {z_1} z_2 + z_1 \overline {z_2} } 2 + \frac {\overline {z_1} z_2 - z_1 \overline {z_2} } 2\) | Definition 4 of Dot Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\overline {z_1} z_2 + z_1 \overline {z_2} + \overline {z_1} z_2 - z_1 \overline {z_2} } 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \overline {z_1} z_2\) |
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Dot and Cross Product