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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 ?

Aditi
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3 Answers3

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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.

dxiv
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$z$ 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 $x=\pm y$.

6

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.

user
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    Could you please explain where I was wrong ? Thanks ! – Aditi Feb 04 '18 at 07:21
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    @Aditi the condition $|z+\bar z|=|z-\bar z|$ doesn't mean that the distance is equal – user Feb 04 '18 at 07:26
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    oh Right !! Thank you – Aditi Feb 04 '18 at 07:26
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    @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 04 '18 at 07:33
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    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):)) – Peter Szilas Feb 04 '18 at 07:36
  • @dxiv Yes you are right of course! I mean that it doaesn't mean that z is equidistant from $\bar z$ and $-\bar z$ as stated. – user Feb 04 '18 at 07:38
  • @gimusi But $,|z+\bar z| = |z-\bar z|,$ *does* mean that $,z,$ is equidistant from $,-\bar z,$ and $,\bar z,$. – dxiv Feb 04 '18 at 07:41
  • @dxiv ops of course I don't know what I was thinking about! – user Feb 04 '18 at 07:45
  • @dxiv actually $|z-\bar{z}|$ represents the length of the resultant of the vectors $z$ and $-\bar{z}$ isn’t it ? I don’t think it represents the distance between them – Aditi Feb 04 '18 at 07:51
  • @Aditi sorry I get confused dxiv if of course right, $|z-\bar z|$ is the distance between $z$ and $\bar z$ and $|z+\bar z|$ is the distance between $z$ and $-\bar z$ and they are equal only if z belongs to the axes bisectors. – user Feb 04 '18 at 07:57
  • @Aditi Of course $,z-\bar z,$ represents the distance between $,z,$ and $,\bar z,$. In general, $,|a-b|,$ represents the distance between $,a,$ and $,b,$, just as $,|c|,$ represents the distance between $,c,$ and $,0,$. – dxiv Feb 04 '18 at 07:57
  • @dxiv oh thank you ! But the prerequisite for it to represent the distance is that $\bar{z}$ should be fixed right ? – Aditi Feb 04 '18 at 07:59
  • @Aditi There is no prerequsite for it. $,|a-b|,$ always represents the distance between $,a,$ and $,b,$. What you derive from that, however, depends on your other assumptions. In the case of $,a=z,$ and $,b=\bar z,$, neither point is fixed, so you can't assume anything about their perpendicular bisector. – dxiv Feb 04 '18 at 08:02
  • @dxiv in general , would , say $|z-2|$ represent the distance between $z$ and $2$ since $2$ is fixed ? I thought it represented the resultant of the vector $z$ and $-2$ just like $|\vec{r_1} - \vec{r_2}|$ represents the length of theresultant of $\vec{r_1}$ And -$\vec{r_2}$ – Aditi Feb 04 '18 at 08:03
  • @Aditi the resultant of the vector z and −2 That's the same as the diagonal between $,z,$ and $,2,$ if you draw the parallelograms. Just as $,\overrightarrow{OA}-\overrightarrow{OB}=\overrightarrow{BA},$ in vector notation. – dxiv Feb 04 '18 at 08:10
  • @dxiv okay thank you very much for clearing my doubt !! – Aditi Feb 04 '18 at 08:19