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

\(\displaystyle \cmod {z - 1}\) \(=\) \(\displaystyle \cmod {z + 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \cmod {z - 1}^2\) \(=\) \(\displaystyle \cmod {z + 1}^2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {z - 1} \paren {\overline {z - 1} }\) \(=\) \(\displaystyle \paren {z + 1} \paren {\overline {z + 1} }\) Modulus in Terms of Conjugate
\(\displaystyle \leadsto \ \ \) \(\displaystyle z \overline z - z - \overline z + 1\) \(=\) \(\displaystyle z \overline z + z + \overline z + 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 2 \paren {z + \overline z}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 4 \map \Re z\) \(=\) \(\displaystyle 0\) Sum of Complex Number with Conjugate
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \Re z\) \(=\) \(\displaystyle 0\)

The result follows by definition of imaginary axis.

$\blacksquare$


Sources