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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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For $z=0$ it is not correct! Otherwise, since $|z|$ is positive, you may cancel it. Then you want to prove that the real part of a complex number is less than the modulus. It holds since one side in a right triangle is always less than the hypotenuse. –  PAD Oct 4 '12 at 11:41
    
I edited my solution per lhf's nice comment. –  Patrick Li Oct 4 '12 at 12:06
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1 Answer

up vote 4 down vote accepted

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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No need to expand into $x$ and $y$: $|z^2|-|z|\ \Re(z)>0$ iff $|z^2|>|z|\ \Re(z)$ iff $|z|>\Re(z)$ iff $\Im(z)\neq0$. –  lhf Oct 4 '12 at 11:41
1  
Not quite: $\sqrt{x^2+y^2} > x$ iff $y\neq0$ or $x<0$. –  lhf Oct 4 '12 at 11:58
    
@lhf You have good eyes. Thanks. –  Patrick Li Oct 4 '12 at 12:07
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