Equation of Line in Complex Plane/Formulation 2

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Theorem

Let $\C$ be the complex plane.

Let $L$ be the infinite straight line in $\C$ which is the locus of the equation:

$l x + m y = 1$


Then $L$ may be written as:

$\map \Re {a z} = 1$

where $a$ is the point in $\C$ defined as:

$a = l - i m$


Proof

Let $z = x + i y$.

Let $a = l - i m$.


Then:

\(\ds \map \Re {a z}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\paren {a z + \overline {a z} } } 2\) \(=\) \(\ds 1\) Sum of Complex Number with Conjugate
\(\ds \leadsto \ \ \) \(\ds a z + \overline a \cdot \overline z\) \(=\) \(\ds 2\) Complex Modulus of Product of Complex Numbers
\(\ds \leadsto \ \ \) \(\ds \paren {l - i m} \paren {x + i y} + \paren {l + i m} \paren {x - i y}\) \(=\) \(\ds 2\) Definition of Complex Conjugate
\(\ds \leadsto \ \ \) \(\ds \paren {\paren {l x + m y} + i \paren {l y - m x} } + \paren {\paren {l x + m y} - i \paren {l y - m x} }\) \(=\) \(\ds 2\) Definition of Complex Multiplication
\(\ds \leadsto \ \ \) \(\ds l x + m y\) \(=\) \(\ds 1\) simplifying

$\blacksquare$


Sources