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Describe the set whose points satisfy the relations $$|{z-1\over z+1}|=1$$ for any $z=a+ib\in\mathbb{C}$.

Solution: $$ \\ 1=|{z-1\over z+1}|=|{(z-1)(z^*+1)\over (z+1)(z^*+1)}| \Rightarrow \\|z+1|^2=|(a^2+b^2-1)+i(2b)|$$

But here I get a complex equation of degree 4 which I don't know to solve.

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  • $\begingroup$ You have two answers which say what I was going to say. $|z-a|$ is the distance from $z$ to $a$, and this is the key to understanding what is going on with so many questions of this kind. $\endgroup$ – Mark Bennet Mar 11 at 22:22
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Remember that for all numbers $z,w\ne 0$ we have $$\Big|{z\over w}\Big| ={|z|\over |w|}$$

So we have $$|z-1| = |z+1|$$ and thus $z$ is at equal distance from $1$ and $-1$ and thus it is perpendicular bisector for segment between $1$ and $-1$, so it is an $y$-axis i.e. $z=bi$ where $b$ is real.

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  • $\begingroup$ I'm not sure of it, because $|{z-1\over z+1}|=1$ is not equivalent to ${|z-1|\over |z+1|}=1$. $\endgroup$ – J. Doe Mar 11 at 22:23
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    $\begingroup$ Why not? ............. can you give an example? $\endgroup$ – Maria Mazur Mar 11 at 22:24
  • $\begingroup$ Ok, that was my mistake, I thought $|z\cdot w| \ne |z|\cdot |w|$. $\endgroup$ – J. Doe Mar 11 at 22:26
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Hint

Think geometrically, $|z-1|=|z+1|$ means all those points which are equidistant from both $1$ and $-1$. So it will be all the points lying on the perpendicular bisector of...

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Just to show your approach using $z=a+bi$ could work:

If $$\left| \frac{z-1}{z+1} \right| = 1 $$ then $|z-1|^2=|z+1|^2, $ so $(a-1)^2+b^2=(a+1)^2+b^2,$ so $(a-1)^2=(a+1)^2.$

Clearly we can't have $a-1=a+1$, so we must have $a-1=-(a+1),$ i.e., $a=0, $

and there are no restrictions on $b$, so the solutions are $a=0+bi$ for all $b \in \mathbb R.$

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  • $\begingroup$ Just to verify, $\left| \frac{bi-1}{bi+1} \right| = \left| \frac {\sqrt{1+b^2}} {\sqrt{1+b^2} }\right| =1 $ $\endgroup$ – J. W. Tanner Mar 11 at 22:47
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$$\left| \frac{z-1}{z+1} \right| = 1 \quad \Leftrightarrow \quad \exists \theta \in \mathbb{R}, \text{} \frac{z-1}{z+1} =e^{i\theta}$$

You deduce that $z(1-e^{i\theta})=e^{i\theta}+1$, i.e. because $e^{i\theta}=1$ can't be a solution, $$z= \frac{1+e^{i\theta}}{1-e^{i\theta}}= i \mathrm{cotan} \left( \frac{\theta}{2} \right)$$

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