Locus of the points on complex plane

Solution 1:

Hint: squaring the inequality, using $\;|w|^2=w \bar w\;$ and regrouping:

$$ \begin{align} 4 z \bar z & \;\gt\; (1+z^2)(1+\bar z ^2) \\ z^2 + \bar z^2 - 2 z \bar z + z^2 \bar z^2 - 2 z \bar z + 1 & \;\lt\; 0 \\ (z-\bar z)^2 + (z \bar z - 1)^2 & \;\lt\; 0 \\ (z \bar z - 1)^2 - \left( i (z-\bar z)\right)^2 & \;\lt\; 0 \\ \left(z \bar z - 1 -i(z-\bar z)\right) \left(z \bar z - 1 + i(z-\bar z)\right) & \;\lt\; 0 \\ \left((z+i)(\bar z-i) - 2\right) \left((z-i)(\bar z +i)-2\right) & \;\lt\; 0 \\ \left(|z+i|^2 - 2\right) \left(|z-i|^2-2\right) & \;\lt\; 0 \end{align} $$

Solution 2:

We are referring to a region outside the intersection, inside union of two separately defined circle boundaries A,B.

$$ A \cap B $$

$$ A: x^2 +(y+1)^2 =2,\, B: x^2 +(y-1)^2 =2\, $$