16

Write the equation of an ellipse, hyperbola, parabola in complex form.

For an ellipse, there are two foci $a,b$, and the sum of the distances to both foci is constant. So $|z-a|+|z-b|=c$.

For a hyperbola, there are two foci $a,b$, and the absolute value of the difference of the distances to both foci is constant. So $||z-a|-|z-b||=c$.

For a parabola, there is a focus $a$ and a line $b+ct$ (where $b,c$ are complex and the parameter $t$ is real.) The distances to both must be equal. The distance to the focus is $|z-a|$. How can we calculate the distance to the line $b+ct$?

Paul S.
  • 3,345
  • 4
  • 32
  • 60
  • Not an easy expression IIRC, but you can get the distance to the line by minimising the square of the distance $|z - b - ct|^2$ using calculus. You may rescale and consider $|c|=1$ to simplify things. – Macavity Sep 01 '13 at 20:28
  • Ahlfors page 17, problem 2. – Supriyo Dec 24 '13 at 17:12

3 Answers3

6

The distance of the point $z$ from the line $b + ct,\; t \in \mathbb{R}$ is the length of the projection of $z-b$ to the normal, which has direction $\pm ic$. If we identify $\mathbb{C}$ with $\mathbb{R}^2$, we'd find the length of the projection by computing the inner product. We do the same in $\mathbb{C}$ even if we don't explicitly identify it with $\mathbb{R}^2$, the real inner product of $v$ and $w$, expressed in complex form, is $\Re \overline{v}w$.

So we get

$$\left\lvert\Re \left(\frac{\overline{ic}(z-b)}{\lvert c\rvert}\right)\right\rvert = \left\lvert\Im \frac{\overline{c}(z-b)}{\lvert c\rvert} \right\rvert$$

as the expression for the distance of $z$ from the line $b + ct$. If $c$ is chosen with absolute value $1$, that simplifies to $\lvert \Im \overline{c}(z-b)\rvert$.

Daniel Fischer
  • 206,697
3

Based on the equations of analytical geometry, one can redefine the equations of a graph in terms of complex variable treating complex plane as two-dimensional space. Consider the equations of parabola in analytical geometry are in the following forms below,

Equation form 1: $$ (y-b)^2= 4 a x $$
Equation form 2:$$ (x-b)^2= 4 a y $$
Let z be a complex variable in a complex plane $\omega$, it is denoted by the following equation $$ z = x + i y $$ where x and y are real and imaginary parts of a complex variable which corresponds to Abscissa and Ordinate in analytical geometry and its conjugate $$\overline{z}= x - i y$$
C be a constant complex number denoted by $$ C= a+i b $$ and its conjugate $$\overline{C}=a- i b$$ Complex representation of Parabola of equation form 1: $$ \mid z -C\mid =\frac{ \mid z + \overline{z}\mid}{2} + \frac{\mid C + \overline{C}\mid}{2} $$ Parabola of equation form 2:$$ \mid z -C\mid = \frac{\mid z - \overline{z}\mid}{2} + \frac{\mid C - \overline{C}\mid}{2} $$

selva kumar
  • 125
1

For an ellipse, there are two foci $a,b$ and the sum of the distances to both foci is constant. So $$|z−a|+|z−b|=c$$

For a hyperbola, there are two foci $a,b$ and the absolute value of the difference of the distances to both foci is constant. So $$||z−a|−|z−b||=c$$

SarGe
  • 3,010
HIMANSHU NAVY SSR
  • 11