inequality - How to solve $|z^2-1|

How to solve $|z^2-1|<|z|^2$ where $z$ is a complex number? I have tried it both with cartesian and polar coordinates but did not get a solution.

I got that far: $z=x+yi$ and then I got: $$\pm x >(\frac{y^2+0.5}{1+4y^2})^{0.5}$$ but I don't know how to visualise that in the coordinate system.

Answer

That is equivalent to
$$|z^2-1|<|z^2|$$
This means that $z^2$ is at a shorter distance from $1$ than from $0$. Then $Re(z^2)>1/2$.

Now, write $z=x+iy$, thus, $z^2=(x^2-y^2)+2xyi$. The former inequality becomes
$$x^2-y^2>\frac12$$