If $|z+\bar{z}| =|z-\bar{z}|$ then determine the locus of $z$.
This is how I attempted it ,
The given statement implies that $z$ is equidistant from -$\bar{z}$ and $\bar{z}$ so it lies on the perpendicular bisector of $z$ and $\bar{z}$ which is a straight line.
However the solution of the given problem is as follows –
Let $z= x+iy$
$|z+\bar{z}| = |z-\bar{z}|$
Implies
$|2x| = |2y|$
$|x|=|y|$
Therefore $x=y$ or $x= -y$ which is a pair of straight lines.
Where did I go wrong ? Is it because a the definition of a perpendicular bisector is the locus of all those points which are equidistant from two fixed points ? But for a given $z$ , wouldn’t $\bar{z}$ and $-\bar{z}$ be fixed ?
Best Answer
But $\,z\,$ is the variable in this case, so the perpendicular bisector of $\,\bar z\,$ and $\,-\bar z\,$ changes for each $\,z\,$.
Instead, use that $\,z+ \bar z = 2 \operatorname{Re}(z)\,$ and $\,z- \bar z = 2i \operatorname{Im}(z)\,$, then the equality reduces to:
$$\require{cancel} |\operatorname{Re}(z)|=|\operatorname{Im}(z)| $$
The latter is equivalent to $\,\operatorname{Re}(z)=\pm \operatorname{Im}(z)\,$, or $\,x=\pm y\,$ in Cartesian coordinates.