Determining locus of a complex number 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 ?
 A: $z$ is not a constant in this problem...  thus neither are $\bar z$ and $-\bar z$...
That's where your error is.
For $z$ to satisfy the equation $\lvert z+\bar z\rvert=\lvert z-\bar z\rvert$, we need only $x=\pm y$.   
A: Note that
$$|z+\bar z|=|2Re(z)|$$
$$|z-\bar z|=|2Im(z)|$$
thus the locus $$|2x|=|2y|\iff|x|=|y|$$
is correct.
Indeed we can also write
$\left|\frac{z+\bar z}2\right|=\left|z-\frac{z-\bar z}2\right| \equiv$ distance of z from y axis
$\left|\frac{z-\bar z}2\right|=\left|z-\frac{z+\bar z}2\right|\equiv$ distance of z from x axis
thus the locus we are looking for are the bisectors.
A: 
The given statement implies that $z$ is equidistant from $-\bar{z}$ and $\bar{z}$ so it lies on the perpendicular bisector of $\ldots$

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.
