<s>[INST] 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?
 [/INST] let $D = \{z \colon 2|z| < |1 + z^2|\}.$ observe that the region $D$ is symmetric with respect to the transformations $z \to -z, z \to \bar z$ and $z \to \dfrac{1}{z}$ therefore it is enough to worry about the portion of $D$ in the first quadrant.  
Let $o = 0, a = 1$ and $z = re^{it}, p = 1 + r^2e^{2it}$ so that $z$ is on the boundary of $D.$ applying rule of cosine to the triangle, we have $$\cos 2t = 2 - \frac{1}{2}\left( r^2 + \frac{1}{r^2} \right).$$  
looking at the graph of $y = 2 - \frac{1}{2}(x + 1/x),$ you can verify that $-1 \le y \le 1$ for $3 - 2\sqrt 2 \le x \le 3 + 2\sqrt 2$ and $f(1) = 1, f(2 \pm \sqrt 3) = 0$ 
i plotted the region $D$ it contains the unit circle and has a hole(lens) containing the origin.
so part of $D$ in the first quadrant is given by 
$$\{re^{it} \colon t = \frac{1}{2}\cos^{-1}\left(  2 - \frac{1}{2}( r^2 + \frac{1}{r^2} ) \right), \sqrt{3 - 2\sqrt 2} \le r \le \sqrt{3 + 2 \sqrt 2} \}.$$ 
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