first post so apologies in advance :)... Disclosure: this is an assignment question I've been given, high school Maths C/first year university Maths.

Solve the simultaneous equations:

$|z+2|=|\bar z-3|$,

for $z \in \Bbb C$.

I've tried solving the first equation substituting $a + bi$ for $z$ and squaring the moduli:

$4a = 0$

so that $a = 0$ and $b$ could be any real number, but this obviously doesn't work in the second equation. So I rearranged both equation to equal zero:

$|z+2|^2-|\bar z-3|^2=|z+1|^2-|z-1|^2$ or
$|z+2|^2+|z-1|^2=|z+1|^2+|\bar z-1|^2$

which fully expanded led me to the solution

$3z + 3\bar z - 5 = 0 \implies \frac{z+\bar z}{2} = a = \frac{-5}{6}$

which $\frac{-5}{6}$ for $a$ does not satisfy equation 1, unless I've made an arithmetical error, though I've checked my working multiple times). Plus, I couldn't see any way to rearrange for $\frac{z-\bar z}{2}$ to get a value for $b$. Another post helped me see that $|z-1|$ is geometrically a circle centred on 1 with a radius thus of $|z+1|$, but I'm having a hard time figuring out what, if anything, to do with that information.

Cutting it short, I'm essentially stuck at either thinking there is no solution that satisfies the equations or (perhaps the more likely option) I'm missing something really simple that I can do to break the deadlock I'm at. Wolfram Alpha gives me a 'numerical solution' of 1.5, but I can't make sense of that answer as a simple substitution doesn't work. Similar questions in the textbook have always ended up forming a quadratic, and usually have a co-efficient in front of one of the moduli. Goes without saying I don't want an explicit solution, I just want to understand where I'm falling short, so I was hoping for a little guidance. Is there a quadratic, for example, that I'm meant to find? Recast a moduli in different terms? Or try to end up with a co-efficient in front of a moduli? Should I be trying to use some additional vector techniques, considering polar forms, or something else...?

Honestly, it's doing my head in. So thanks in advance :)

Thanks everyone for your help and rapid responses, I sincerely appreciate it!


3 Answers 3


\begin{array}{c} |z+1|=|z-1| \\ |z+2|=|\bar z-3| \end{array}

Since $|\bar z-3| = |z-3|$ we can rewrite this as

\begin{array}{c} |z+1|=|z-1| \\ |z+2|=|z-3| \end{array}

In terms of the distance function, $d:\mathbb C \times \mathbb C \to \mathbb R$

\begin{array}{c} d(z,-1)=d(z,1) \\ d(z-2)=d(z,3) \end{array}

In general, $d(z,a)=d(z,b)$ describes the perpendicular bisector of the segment with endpoints at $a$ and $b$. If $a$ and $b$ are real numbers, this describes the vertical line $z = \frac 12(a+b)+ti$, where $t \in \mathbb R$

Since the lines $z=0 + ti$ and $z=\frac 12 + ti$ are parallel, there is no solution.


You know already that $a=0$ so You only have to consider the second equation with $z=bi$. This system has no solution, by the way.


So, the second equation gives



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