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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

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    I edited my solution per lhf's nice comment.2012-10-04

1 Answers 1

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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 $z=\{x+iy: y\neq 0\ \text{ or } x <0\}$.

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    @lhf You have good eyes. Thanks.2012-10-04