We have a complex plane and there are inequality on it: $2|z| > |1 + z^2|$. What is the locus of such set of $z$?
I tried to solve it by using polar coordinates and got: $r^4 - 2r^2\cos(2\phi ) - 1 = 0$.
As well I tried solve it using $z = x + iy$ and got: $4(x^2 + y^2) = (x^2 - y^2 + 1)^2 + 4x^2y^2$
By I still can`t figure out what is the explicit name or at least descritpion for such locus of points. Will be very grateful for any help.
Hint: squaring the inequality, using $\;|w|^2=w \bar w\;$ and regrouping:
$$ \begin{align} 4 z \bar z & \;\gt\; (1+z^2)(1+\bar z ^2) \\ z^2 + \bar z^2 - 2 z \bar z + z^2 \bar z^2 - 2 z \bar z + 1 & \;\lt\; 0 \\ (z-\bar z)^2 + (z \bar z - 1)^2 & \;\lt\; 0 \\ (z \bar z - 1)^2 - \left( i (z-\bar z)\right)^2 & \;\lt\; 0 \\ \left(z \bar z - 1 -i(z-\bar z)\right) \left(z \bar z - 1 + i(z-\bar z)\right) & \;\lt\; 0 \\ \left((z+i)(\bar z-i) - 2\right) \left((z-i)(\bar z +i)-2\right) & \;\lt\; 0 \\ \left(|z+i|^2 - 2\right) \left(|z-i|^2-2\right) & \;\lt\; 0 \end{align} $$
We are referring to a region outside the intersection, inside union of two separately defined circle boundaries A,B.
$$ A \cap B $$
$$ A: x^2 +(y+1)^2 =2,\, B: x^2 +(y-1)^2 =2\, $$
