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

Question

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.

Answer 1

The distance from $z$ to $0$ is the same as the distance from $1$, so it has to be on the line that is perpendicular to the segment connecting $0$ to $1$, i.e.

$$\text{Re}(z)=\text{Re}(x+iy)=x={1\over 2}$$

Then you need that it also has

$$|z|=|z^{-1}|=|z|^{-1}\iff |z|^2=1$$

(just multiply both sides by $|z|$ if you want to see this)

This means that $|z|^2=x^2+y^2=\left({1\over 2}\right)^2+y^2=1$

which you can easily solve to see $y=\pm{\sqrt{3}\over 2}$.

Answer 2

If $z=re^{i2\theta}$ where $r\ge0,\theta$ are real

$|z^{-1}|=\dfrac1r\implies r=\dfrac1r\iff r^2=1\implies r=1$

$z-1=e^{i2\theta}-1=\cos2\theta+i\sin2\theta-1=2i\sin\theta(\cos\theta+i\sin\theta)$

$\implies |z-1|=2|\sin\theta|$

$|z|=|z-1|\implies 2|\sin\theta|=1\iff\sin\theta=\pm\frac12\implies\cos2\theta=1-2\sin^2\theta=\dfrac12$