Product of Complex Number with Conjugate in Exponential Form
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Theorem
Let $z_1$ and $z_2$ be complex numbers.
Then:
- $\overline {z_1} z_2 = \cmod {z_1} \, \cmod {z_2} e^{i \theta}$
where:
- $\overline {z_1}$ denotes the complex conjugate of $z_1$
- $\cmod {z_1}$ denotes the complex modulus of $z_1$
- $\theta$ denotes the angle from $z_1$ to $z_2$, measured in the positive direction.
Proof
\(\ds \overline {z_1} z_2\) | \(=\) | \(\ds \paren {z_1 \circ z_2} + i \paren {z_1 \times z_2}\) | Product of Complex Number with Conjugate by Dot and Cross Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {z_1} \, \cmod {z_2} \cos \theta + i \paren {z_1 \times z_2}\) | Definition 2 of Dot Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {z_1} \, \cmod {z_2} \cos \theta + i \cmod {z_1} \, \cmod {z_2} \sin \theta\) | Definition 2 of Vector Cross Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {z_1} \, \cmod {z_2} \paren {\cos \theta + i \sin \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {z_1} \, \cmod {z_2} e^{i \theta}\) | Euler's Formula |
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Dot and Cross Product