Suppose $u,v$ are points, describe the set of $z$ satisfy $|z-u|=|z-v|$ Suppose $u,v$ are points, describe the set of $z$ satisfy $|z-u|=|z-v|$
My attempt
Let $z=x+iy$
Then
$|x+iy-u|=|x+iy-v|\iff (x-u)^2+(iy)^2=(x-v)^2+(iy)^2\iff 2ux+u^2=2vx+v^2\iff 2x(u-v)+u^2-v^2=0\iff u=v$ 
Then this set is:
$S=\{z\in \mathbb{C}:|z-t|=|z-t|,t \in \mathbb{R} \}$
Is correct this? How i can interpret this geometrically?
 A: Fix $u=a+bi$ and $v=a'+b'i$. We are looking for the set $S$ of complex numbers such that $z=x+yi\in S$ if and only if the distance from $z$ to each of $u$ and $v$ is the same. It's easy to see by drawing a picture, that the graph of $S$ is a line in the complex plane bisecting the segment between $u$ and $v$. 
Now to find the equation of this line, we will exploit the fact that $\mathbb C\cong \mathbb R^2$ and do a little analytic geometry in the plane: 
The segment $\overline {uv}$ is $\overline {uv}=\left \{ (1-t)u+tv:0\le t\le 1 \right \}$ and the midpoint is $\frac{1}{2}(a+a',b+b').$ A vector parallel to the segment is $\vec y=(a'-a,b'-b)$, and so a vector perpendicular to $\vec y$ is $\vec y_p=(b-b',a'-a).$ 
So, we have what we want: a line in direction of $\vec y_p$ passing through $\frac{1}{2}(a+a',b+b')$, and so the parametric equation of the line is $(x,y)=\frac{1}{2}(a+a',b+b')+t(b-b',a'-a);\ t\in \mathbb R.$ 
A: Note that $u$ and $v$ are not necessarily real numbers so we have to consider the case where they are complex numbers.
Since $|z-u|$ is the distance between $z$ and $u$ and  $|z-v|$ is the distance between $z$ and $v$, The equation $$ |z-u| =|z-v|$$ is equivalent to the statement that $z$ is equidistant from  $u$ and $v$
That is to say, $z$ in on the perpendicular bisector of the segment joining $u$ and $v$
Squaring both sides, we get $$ |z-u|^2 =|z-v|^2$$Which could be expressed as $$(z-u)\bar {(z-u)} = (z-v)\bar {(z-v)}$$
