# Geometric meaning of | |z - z_1| - |z -

Geometric meaning of $||z-{z}_{1}|-|z-{z}_{2}\mid \mid =a$, where $z,{z}_{1},{z}_{2}\in \mathbb{C}$
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smachttenbem
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 $P{F}_{1},P{F}_{2}$ 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.
$||z-{z}_{2}|-|z-{z}_{1}\mid \mid =c$
$||\left(x+iy\right)-\left({x}_{2}+i{y}_{2}\right)|-|\left(x+iy\right)-\left({x}_{1}+i{y}_{1}\right)\mid \mid =c$
$||\left(x-{x}_{2}\right)+i\left(y-{y}_{2}\right)|-|\left(x-{x}_{1}\right)+i\left(y-{y}_{1}\right)\mid \mid =c$
$|\sqrt{{\left(x-{x}_{2}\right)}^{2}+{\left(y-{y}_{2}\right)}^{2}}-\sqrt{{\left(x-{x}_{1}\right)}^{2}+{\left(y-{y}_{1}\right)}^{2}}|=c$
$\sqrt{{\left(x-{x}_{2}\right)}^{2}+{\left(y-{y}_{2}\right)}^{2}}-\sqrt{{\left(x-{x}_{1}\right)}^{2}+{\left(y-{y}_{1}\right)}^{2}}=±c$
Now, if you have an equation of the form $\sqrt{u}-\sqrt{v}=w$, then doing some algebra gives you ${\left({w}^{2}-u-v\right)}^{2}=4uv$, eliminating the inconvenient √ signs.
${\left({c}^{2}-{\left(x-{x}_{2}\right)}^{2}-{\left(y-{y}_{2}\right)}^{2}-{\left(x-{x}_{1}\right)}^{2}-{\left(y-{y}_{1}\right)}^{2}\right)}^{2}=4\left({\left(x-{x}_{2}\right)}^{2}+{\left(y-{y}_{2}\right)}^{2}\right)\left({\left(x-{x}_{1}\right)}^{2}+{\left(y-{y}_{1}\right)}^{2}\right)$