Find all real numbers $a$ for which there exists a complex number $z$ such that $|z|=1$ and $|az-1|=a|z+1|$

Question

Find all real numbers $a$ for which there exists a complex number $z$ such that $|z|=1$ and $|az-1|=a|z+1|$.

An effort: I first note that for $z=-1$ we have $a=-1$ as well. Now, let $z+1$ not be equal to $0$. I divide both sides in $|az-1|=a|z+1|$ by $|z+1|$ and I'm lost.

Answer 1

$a$ has to be non-negative real number, the only exceptional case being when $az-1=0$. In this case we get $a=-1$ and $z =-1$. Now consider the case $a \geq 0$. The second condition says $a^{2}|z|^{2}+1-2a\Re z =a^{2}|z|^{2}+a^{2}+2a^{2}\Re z$ which becomes $2(a+a^{2})\Re z=1-a^{2}$. So a complex number $z$ satisfying the two conditions exists iff $a \geq 0$ and $\frac {|1-a^{2}|} {2(a+a^{2})} \leq 1$ or $|1-a^{2}| \leq 2|a||1+a|$.

I leave it to you to check that this is true iff $a=-1$ or $a \geq \frac 1 3$