Solving the equation $|z^2-z+1|=|x^2-1|$

Question

$z \in \mathbb{U}$, let $x= |z-1|$, show that :$|z^2-z+1|=|x^2-1|$

I tried from both sides but nothing worked for me, any ideas ?

Answer 1

Well, $x^2=(z-1)(\bar{z}-1)=z\bar{z}-z-\bar{z}+1$. Since $\bar{z}=\dfrac{1}{z}$ for $z\in\mathbb{U}$, we get $$x^2-1=\frac{z-z^2-1}{z}=-\frac{z^2-z+1}{z}\,.$$ That is, $$\left|x^2-1\right|=\frac{\left|z^2-z+1\right|}{|z|}=\left|z^2-z+1\right|\,.$$

Answer 2

Hint: With $$z=x+iy$$ we get $$|x^2-y^2+2xyi-x-iy+1|=|(x-1)^2+y^2-1|$$ Can you proceed?