Cline

Notebook post 




this article expresses complex conjugate of \(c\) with asterisk notation \(c^*\)



Implicit | Clines


the set of all real numbers \(\mathbb{R}\) forms a horizontal axis on complex plane. Its element satisfy

$$r=r^* \quad \quad \dfrac{r}{r^*}=1$$

the set of all unit complex number (denoted \(\mathbb{T}\) or \(\mathbb{S}^1\) forms a unit circle on complex plane. Its elements satisfy

$$s s^* = 1$$

complex number \(z\) that lies on a perpendicular bisector between \(0\) and \(1\) satisfy

$$z + z^* = 1$$

complex number \(z\) that lies on orthogonal axis satisfy

$$z + z^* = 0$$

complex number \(z\) that lies on a line across \(a_0\) and \(a_1\) satisfy

$$(z - a_0)(a_1 - a_0)^* = (z - a_0)^*(a_1 - a_0)$$

$$z(a_1 - a_0)^* -z^*(a_1 - a_0) = a_0 a_1 ^* - a_0 ^* a_1$$

complex number \(z\) that lies on a circle centered at \(a_0\) across \(a_1\) satisfy

$$(z - a_0)(z - a_0)^* = (a_1 - a_0)(a_1 - a_0)^*$$

$$zz^* - a_0 ^* z - a_0 z^* = a_1 a_1 ^* - a_1 a_0^* - a_1 ^* a_0 $$

complex number \(z\) that lies on a perpendicular bisector between \(a_0\) and \(a_1\) satisfy

$$z(a_1 - a_0)^* + z^*(a_1 - a_0) = a_1 a_1 ^* - a_0 a_0 ^*$$

in inversive geometry, lines and circles are indistinguishable. For completeness, a complex number \(z\) that lies on a line or circle across \(a_0\) ,\(a_1\) , and \(a_2\) satisfy

$$(z - a_1)(z - a_2)^*(a_0 - a_2)(a_0 - a_1)^* = (z - a_1)^*(z - a_2)(a_0 - a_2)^*(a_0 - a_1)$$

or

$$p_0 zz^* + p_1 z + p_2 z^* + p_3 = 0$$

where

$$\begin{array}{llll} p_0 &= a_0^* a_1 - a_0 a_1^* + a_1^* a_2 - a_1 a_2^* + a_2^* a_0 - a_2 a_0^* \\ \\ p_1 &= a_0 a_0^* (a_1 - a_2)^* + a_1 a_1^* (a_2 - a_0)^* + a_2 a_2^* (a_0 - a_1)^*\\ \\ p_2 &= -p_1^* \\ \\ &= a_0 a_0^* (a_2 - a_1) + a_1 a_1^* (a_0 - a_2) + a_2 a_2^* (a_1 - a_0)\\ \\ p_3 &= a_0 a_0^* (a_1 a_2^* - a_1^* a_2) + a_1 a_1^* (a_2 a_0^* - a_2^* a_0) + a_2 a_2^* (a_0 a_1^* - a_0^* a_1) \end{array}$$

the center is

$$z = \dfrac{(\;a_1 a_1^* - a_0 a_0^*\;)(\;a_2 - a_1\;) - (\;a_2 a_2^* - a_1 a_1^*\;)(\;a_1 - a_0\;)}{(\;a_1 - a_0\;)^*(\;a_2 - a_1\;) - (\;a_2 - a_1\;)^*(\;a_1 - a_0\;)}$$

Under complex inversion :

circles not crossing \(0\) maps into circles not crossing \(0\)

circles across \(0\) maps into lines not crossing \(0\)

lines not crossing \(0\) maps into circles across \(0\)

lines across \(0\) maps into lines across \(0\)

a line across \(a_0\) and \(a_1\) is a perpendicular bisector between \(0\) and 

$$c^{-1}=\dfrac{a_0 a_1 ^* - a_0 ^* a_1}{(a_1 - a_0)^*}$$

under inversion, this line is mapped into circle centered at

$$c = \dfrac{(a_1 - a_0)^*}{a_0 a_1 ^* - a_0 ^* a_1}$$

a circle centered at \(a_0\) across \(a_1\) not crossing \(0\) under inversion is mapped into another circle centered at 

$$c=\dfrac{a_0^*}{a_1 a_0^* +a_1 ^*a_0 -a_1 a_1 ^*}$$

if the circle does cross point of origin \(0\), under inversion is mapped into a perpendicular bisector between \(0\) and

$$c=\dfrac{1}{a_0}$$

Miscallenous


reflection of a point \(a_2\) over a line across \(a_0\) and \(a_1\) is 

$$a_3 = (a_1 - a_0)\left(\dfrac{a_2-a_0}{a_1-a_0}\right)^*+a_0$$

a line across point \(a_2\) perpendicular to a line across \(a_0\) and \(a_1\) is

$$z\left((a_1 - a_0)\left(\dfrac{a_2-a_0}{a_1-a_0}\right)^*+a_0 - a_2\right)^* -z^*\left((a_1 - a_0)\left(\dfrac{a_2-a_0}{a_1-a_0}\right)^*+a_0 - a_2\right) = a_2 \left((a_1 - a_0)\left(\dfrac{a_2-a_0}{a_1-a_0}\right)^*+a_0\right) ^* - a_2 ^* (a_1 - a_0)\left(\dfrac{a_2-a_0}{a_1-a_0}\right)^*+a_0$$

if \(a_2\) is located on that line, then 

$$z(a_0 - a_2)^*+z^*(a_0 - a_2) = a_0 a_0 ^* - a_2 a_2 ^*$$

a line across point \(a_2\) parallel to a line across \(a_0\) and \(a_1\) is

$$(z - a_2)(a_1-a_0)^*=(z - a_2)^*(a_1-a_0)$$

$$z(a_1 - a_0)^* - z^*(a_1 - a_0) = a_2 (a_1 - a_0)^* - a_2 ^* (a_1 - a_0)$$

Parametric | Clines


line segment from \(0\) to \(1\) can be written as set of numbers \(z\) where

$$z=t \quad,\quad t= [0,1]$$

circle centered at \(0\) across \(1\) can be written as set of numbers \(z\) where

$$z=e^{it}\quad,\quad t= [0,2\pi]$$

however one can also map the entire real number into unit circle with

$$z=\dfrac{-it+1}{it+1}=\dfrac{-(it-1)^2}{t^2+1}$$

line segment from \(a_0\) to \(a_1\), the interpolation is linear

$$z=t\cdot(a_1 - a_0) + a_0 \quad,\quad t= [0,1]$$

circle centered at \(a_0\) across \(a_1\), the interpolation is circular

$$z=e^{it}\cdot(a_1 - a_0) + a_0\quad,\quad t= [0,2\pi]$$






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