Equation of Circle in Complex Plane/Formulation 2/Proof 1

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Theorem

Let $\C$ be the complex plane.

Let $C$ be a circle in $\C$.


Then $C$ may be written as:

$\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$

where:

$\alpha \in \R_{\ne 0}$ is real and non-zero
$\gamma \in \R$ is real
$\beta \in \C$ is complex such that $\cmod \beta^2 > \alpha \gamma$.

The curve $C$ is a straight line if and only if $\alpha = 0$ and $\beta \ne 0$.


Proof

\(\ds \cmod {z - a}\) \(=\) \(\ds r\) Equation of Circle in Complex Plane: Formulation 1
\(\ds \leadsto \ \ \) \(\ds \cmod {z - a}^2\) \(=\) \(\ds r^2\)
\(\ds \leadsto \ \ \) \(\ds \paren {z - a} \overline {\paren {z - a} }\) \(=\) \(\ds r^2\) Modulus in Terms of Conjugate
\(\ds \leadsto \ \ \) \(\ds \paren {z - a} {\paren {\overline z - \overline a} }\) \(=\) \(\ds r^2\) Sum of Complex Conjugates
\(\ds \leadsto \ \ \) \(\ds z \overline z - a z - \overline a \overline z + a \overline a\) \(=\) \(\ds r^2\)
\(\ds \leadsto \ \ \) \(\ds \alpha z \overline z - \alpha a z - \alpha \overline a \overline z + \alpha a \overline a - r^2\) \(=\) \(\ds 0\) where $\alpha \in \R_{>0}$ is arbitrary

By setting:

$\beta := -\alpha a$ and $\gamma = \alpha a \overline a - r^2$

we have:

$\alpha z \overline z + \beta z + \overline \beta \overline z + \gamma = 0$


We have that:

\(\ds r^2\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \alpha a \overline a - \gamma\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {\alpha a} \paren {\alpha \overline a}\) \(>\) \(\ds \alpha \gamma\)
\(\ds \leadsto \ \ \) \(\ds \beta \overline \beta\) \(>\) \(\ds \alpha \gamma\)
\(\ds \leadsto \ \ \) \(\ds \cmod {\beta}^2\) \(>\) \(\ds \alpha \gamma\) Modulus in Terms of Conjugate

$\Box$


If $\alpha = 0$ and $\beta \ne 0$ the equation devolves to:

$\beta z + \overline \beta \overline z + \gamma = 0$

which from Equation of Line in Complex Plane: Formulation 1 is the equation of a straight line.


The result follows.

$\blacksquare$


Sources

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