Equation of Imaginary Axis in Complex Plane
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Theorem
Let $\C$ be the complex plane.
Let $z \in \C$ be subject to the condition:
- $\cmod {z - 1} = \cmod {z + 1}$
where $\cmod {\, \cdot \,}$ denotes complex modulus.
Then the locus of $z$ is the imaginary axis.
Proof
\(\ds \cmod {z - 1}\) | \(=\) | \(\ds \cmod {z + 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \cmod {z - 1}^2\) | \(=\) | \(\ds \cmod {z + 1}^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z - 1} \paren {\overline {z - 1} }\) | \(=\) | \(\ds \paren {z + 1} \paren {\overline {z + 1} }\) | Modulus in Terms of Conjugate | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds z \overline z - z - \overline z + 1\) | \(=\) | \(\ds z \overline z + z + \overline z + 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \paren {z + \overline z}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 4 \map \Re z\) | \(=\) | \(\ds 0\) | Sum of Complex Number with Conjugate | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \Re z\) | \(=\) | \(\ds 0\) |
The result follows by definition of imaginary axis.
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: Example $3$: $(2.5)$