<s>[INST] Describe geometry of complex plane. 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.
 [/INST] 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 $a^2-c^2\ne 0$
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