Geometric meaning of $| |z - z_1| - |z - z_2| | = a$, where $z, z_1, z_2 \in \mathbb C$

Question

Find locus of points of $$| |z - z_2| - |z - z_1| | = c, \ \ c < |z_1 - z_2| $$
$a, z_1, z_2$ are constants
I know, that this is a hyperbola, but I can't show that and bring the equation to canonical form: $$ \frac{Re(z)^2}{a^2} - \frac{Im(z)^2}{b^2} = 1$$
I know, that I can square the equation several times, but I wasted a lot of time, doing this way and didn't get the answer. Is there another way to solve it?

Answer 1

One way to define a hyperbola is

a set of points, such that for any point $P$ of the set, the absolute difference of the distances $PF_1$, $PF_2$ to two fixed points $F_1$, $F_2$ (the foci) is constant

or, using the complex-plane notation from your question,

a set of points, such that for any point $z$ of the set, the absolute difference $c$ of the distances $|z - z_1|$, $|z - z_2|$ to two fixed points $z_1$, $z_2$ (the foci) is constant

So yes, the locus is a hyperbola.

The mistake you're making when you try to “bring the equation to canonical form” $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is that that form assumes that the two foci are on the x-axis and equidistant from the origin, whereas an arbitrary hyperbola can have its foci anywhere.

If you need an explicit equation, it may be helpful to break each complex value into its real ($x$) and imaginary ($y$) components.

$$||z-z_2|−|z-z_1||=c$$ $$||(x+iy)-(x_2+iy_2)|−|(x+iy)-(x_1+iy_1)||=c$$ $$||(x-x_2) + i(y-y_2)|−|(x-x_1) + i(y-y_1)||=c$$ $$| \sqrt{(x-x_2)^2 + (y-y_2)^2} - \sqrt{(x-x_1)^2 + (y-y_1)^2}|=c$$ $$\sqrt{(x-x_2)^2 + (y-y_2)^2} - \sqrt{(x-x_1)^2 + (y-y_1)^2}=±c$$

Now, if you have an equation of the form $\sqrt{u} - \sqrt{v} = w$, then doing some algebra gives you $(w^2 - u -v)^2 = 4uv$, eliminating the inconvenient √ signs.

$$(c^2 -(x-x_2)^2 - (y-y_2)^2 - (x-x_1)^2 - (y-y_1)^2)^2 = 4((x-x_2)^2 + (y-y_2)^2)((x-x_1)^2 + (y-y_1)^2)$$