Find all complex number $z\in\Bbb{C}$ such that $\vert z\vert=\vert z^{-1}\vert=\vert z-1\vert$ 
Find all complex number $z\in\Bbb{C}$ such that $$\vert z\vert=\vert z^{-1}\vert=\vert z-1\vert$$ 

I tried to write $z=a+ib$, clearly $z=1$ is not a solution. I have to solve 
$$\left\{
\begin{array}{l}
  a^2+b^2=1 \\
  1=\sqrt{(a-1)^2-b^2}\sqrt{a^2+b^2}
\end{array}
\right.$$
By multiplying $\sqrt{a^2+b^2}$, (I hope it's correct). 
Which it implies to solve 
$$
\left\{
\begin{array}{l}
  a^2+b^2=1 \\
  1=(a-b)^2\bigl((a^2-b^2)-2a\bigr)
\end{array}
\right.
$$
Here I am stuck, thanks in advance for your help.

If $z=re^{i2\theta}$ where $r\ge0,\theta$ are real
$|z^{-1}|=\dfrac1r\implies r=\dfrac1r\iff r^2=1\implies r=1$
$z-1=e^{i2\theta}-1=\cos2\theta+i\sin2\theta-1=2i\sin\theta(\cos\theta+i\sin\theta)$
$\implies |z-1|=2|\sin\theta|$
$|z|=|z-1|\implies 2|\sin\theta|=1\iff\sin\theta=\pm\frac12\implies\cos2\theta=1-2\sin^2\theta=\dfrac12$
