Describe geometry of complex plane.

Question

Let $a \in \Bbb R$ and $c>0$ to be fixed. Describe the set of points $z$ such that $|z-a|-|z+a|=2c$ for every possible choice of $a$ and $c$.

Then let $a$ be a complex number using the rotation of the plane describe the locus of points satisfying the above equation.

Here is how I understand:

The set of point $z$ is the intersect of $2$ disks, $1$ disk contain all points $z$ that has center $a$ and radius $|z-a|$ and another disk contain all point $z$ that has center $-a$ and radius $|z+a|$.

Since $c>0$, $|z-a|>|z+a|$ that mean the set point $z$ we want to find is closer to $-a$ than $a$. I'm not sure if this is how they want me to describe the set of point $z$.

For the second part, is it just be the same, the only difference is on part $2$, $a$ is on complex plane not a real line.

Answer 1

with $z=x+iy$ it is equivalent to $$\sqrt{(x-a)^2+y^2}-\sqrt{(x+a)^2+y^2}=2c$$ Does this help you? squaring the equation $$\sqrt{(x-a)^2+y^2}=2c+\sqrt{(x+a)^2+y^2}$$ we get $$-xa-c^2=c\sqrt{(x+a)^2+y^2}$$ squaring again we get $$x^2a^2+c^4+2xac^2=c^2((x+a)^2+y^2)$$ this is equivalent to $$x^2(a^2-c^2)+c^2(c^2-a^2)=y^2c^2$$ or $$\frac{x^2}{c^2}-\frac{y^2}{a^2-c^2}=1$$ if $c\ne 0$ and $a^2-c^2\ne 0$

Answer 2

Hint: the points $z \in \Bbb C$ such that the difference of the distance between $z$ and $(a,0)$ and the distance between $z$ and $(-a,0)$ is a constant $2c$ is a hyperbola. The points $(a,0)$ and $(-a,0)$ are the foci of the hyperbola. The sign of $c$ determines how the hyperbola is located in the plane. Identify $\Bbb R^2 \cong \Bbb C$ via $z = x+{\rm i}y \equiv (x,y)$, forget complex numbers and think geometrically.

Answer 3

Since you are interested in an intersection of two vectors better to think of vectors $ z+a $ and $ z-a$ geometrically and imagine circular arcs centered at $ ( \pm c,0)$ intersecting as shown. Difference of distances is a constant,the locus is a hyperbola with a = 1, OB = c.