# 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 ?

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.

$z$ is not a constant in this problem... thus neither are $\bar z$ and $-\bar z$...

For $z$ to satisfy the equation $\lvert z+\bar z\rvert=\lvert z-\bar z\rvert$, we need only $x=\pm y$.

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.

• Could you please explain where I was wrong ? Thanks ! Feb 4 '18 at 7:21
• @Aditi the condition $|z+\bar z|=|z-\bar z|$ doesn't mean that the distance is equal
– user
Feb 4 '18 at 7:26
• oh Right !! Thank you Feb 4 '18 at 7:26
• @gimusi doesn't mean that the distance is equal But yes, it means precisely that for each $\,z\,$, just as $\,|z-a|=|z-b|\,$ means that $\,z\,$ is equidistant from $\,a,b\,$ in general. Problem in OP's case is that $\,a,b\,$ were not fixed points, but varied with $\,z\,$ itself.
– dxiv
Feb 4 '18 at 7:33
• Aditi.For a second I was puzzled by your argument :Equidistant from 2 fixed points. plus minus z (upper bar) are NOT fixed points , as z varies so does z(upper bar):)) Feb 4 '18 at 7:36