# Describe the set whose points satisfy the relation $|{z-1\over z+1}|=1$

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.

• 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. – Mark Bennet Mar 11 at 22:22

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.

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

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...

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 $$b \in \mathbb R.$$

• Just to verify, $\left| \frac{bi-1}{bi+1} \right| = \left| \frac {\sqrt{1+b^2}} {\sqrt{1+b^2} }\right| =1$ – J. W. Tanner Mar 11 at 22:47

$$\left| \frac{z-1}{z+1} \right| = 1 \quad \Leftrightarrow \quad \exists \theta \in \mathbb{R}, \text{} \frac{z-1}{z+1} =e^{i\theta}$$

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