locus of a variable complex number

Question

If $z$ is a variable complex number , and $a$ is a fixed complex number , is it true that if $z$ , $a$ satisfy the following condition

$|z+a| = |z-a|$

Then the locus of $z$ is the perpendicular bisector of $a$ and $-a$ ?

Answer 1

Of course, you have to assume $a\ne0$. Write $a=ru$, with $|u|=1$ and $r>0$; then, writing $z=wu$, the equation becomes $$ |w-r|=|w+r| $$ Notice that this corresponds to a rotation around the origin by the negative of the angle determined by $u$.

By squaring, $$ (w-r)(\bar{w}-r)=(w+r)(\bar{w}+r) $$ that simplifies to $$ w+\bar{w}=0 $$ that is, $w$ is purely imaginary. The locus is therefore the $y$-axis, the perpendicular bisector of the segment joining $-r$ and $r$. Multiplying back by $u$, which is a rotation, we finish.

Answer 2

let $$z=x+iy$$ then $$|x+a+iy|=|x-a+iy|$$ and we get $$\sqrt{(x+a)^2+y^2}=\sqrt{(x-a)^2+y^2}$$ Can you proceed?