<s>[INST] Solve the inequality $|z^2|-|z|\ \Re(z)>0$ Determine the set of complex numbers $z$ such that
$|z|^2-|z|\ \Re(z)>0$
This is my process:
Putting $z=x+iy$, we have $\Re(z)=x$ (real part), $|z^2|=x^2+y^2$, $|z|=\sqrt{x^2+y^2}$, and:
$(x^2+y^2)-x\sqrt{x^2+y^2}>0\Rightarrow x^2\left(1+\frac{y^2}{x^2}\right)-x|x|\sqrt{1+\frac{y^2}{x^2}}>0$
If $x\geq0$ we have $x^2\left(1+\frac{y^2}{x^2}\right)-x^2\sqrt{1+\frac{y^2}{x^2}}>0$, and for $x\neq 0$:
$t-\sqrt{t}>0\ \ \ \ \ \left[t=\left(1+\frac{y^2}{x^2}\right)\right]$ from this: $t>1$ ($t<0$ is not acceptable) i.e. $1+\frac{y^2}{x^2}>1\Rightarrow \frac{y^2}{x^2}>0\Rightarrow y\neq 0$
Then $S_1=\left\{(x,y): x>0, y\neq 0\right\}$
If $x<0$ we have $x^2\left(1+\frac{y^2}{x^2}\right)+x^2\sqrt{1+\frac{y^2}{x^2}}>0$ always verified.
Then $S_2=\left\{(x,y): x<0\right\}$.
What do you think of my proceedings? Is my procedure right? I made ​​a mistake?
thank you very much
 [/INST] I think from $$(x^2+y^2)-x\sqrt{x^2+y^2}>0$$
$$\to (x^2+y^2) > x\sqrt{x^2+y^2}$$
If $x$ and $y$ are not both $0$, divide by $\sqrt{x^2+y^2}$ on both side
$$\sqrt{x^2+y^2} > x$$
which is always true as long as $y\neq 0$ or $x< 0$.
So the solution would be $z=\{x+iy: y\neq 0\ \text{ or } x <0\}$.
</s>