Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$ Determine the set of points $z$ that satisfy the condition $|2z|>|1+z^2|$
I tried to redo this problem and got to this point
$|2z|>|1+z^2|$ $\Rightarrow$ $2|z|>1+|z^2|$ $\Rightarrow$ $2|z|>1+z\overline z$
Let $z=x+iy$ then 
$$2\sqrt{x^2+y^2}>1+(x+iy)(x-iy)$$
$$2\sqrt{x^2+y^2}>1+(x^2+y^2)$$
$$0>1-2\sqrt{x^2+y^2}+(x^2+y^2)$$
$$0>(1-\sqrt{x^2+y^2})^2$$
since $x,y$ are real number , so there is no solution for this inequality?

Let $z=re^{i\theta}$. Then after squaring both sides we have
$$
4r^2 > r^4 + 2r^2\cos(2\theta) + 1\Rightarrow
0>(r^2-2r\sin(\theta)-1)(r^2+2r\sin(\theta)-1)\tag{1}
$$
where I used the following identity $\cos(2\theta) = 1 -2\sin^2(\theta)$. Let's write $(1)$ in Cartesian coordinates so recall that $r^2 = x^2 + y^2$ and $r\sin(\theta) = y$. Then we have
$$
0>(x^2+y^2-2y-1)(x^2+y^2+2y-1)=[x^2+(y-1)^2-2][x^2+(y+1)^2-2]
$$
Then
\begin{align}
x^2+(y-1)^2&< 2\\
x^2+(y+1)^2&< 2
\end{align}
That is, we have two disc of radius $\sqrt{2}$ centered at $(0,\pm i)$. So the solution set is
$$
S=\bigl\{z\in\mathbb{C}\colon \lvert z-i\rvert < \sqrt{2}\text{ or }\lvert z+i\rvert < \sqrt{2}\text{ but not in the intersection}\bigr\}
$$
