find all complex numbers $z$ such that $\mid 2z/(z^2+1) \mid < 1$ I need to find all complex numbers $z$ such that $\mid 2z/(z^2+1) \mid < 1$. If $z$ were real, it would be easy, but now that $z$ being complex numbers makes things so twisted...I cannot find a way to approach. Could anyone please help me?

If you substitute $z=x+iy$ into this inequality then square both sides you should get
$$x^4+y^4+2x^2y^2-2x^2-6y^2+1\gt0$$
$$(x^2+(y-1)^2-2)(x^2+(y+1)^2-2)\gt0$$
So we need
$$z\in\{x+iy:x^2+(y-1)^2-2\gt0\land x^2+(y+1)^2-2\gt0\}\cup\{x+iy:x^2+(y-1)^2-2\lt0\land x^2+(y+1)^2-2\lt0\}$$
