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Let $z\in\mathbb{C}$.

What is the correct way of square rooting both sides of the inequality $$\text{Im}(z)^2 < 3\text{Re}(z)^2\;\text{?}$$

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Both sides of the inequality are real numbers, so you handle them "as usual". –  Austin Mohr Nov 24 '12 at 22:01
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2 Answers

$$ |\text{Im}\,(z)| < \sqrt3\,|\text{Re}\,(z)|. $$

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Are the modulus signs needed on the LHS as $\text{Im}(z) \in \mathbb{R}$? I'm assuming you meant a strictly less than sign? –  Sam Nov 24 '12 at 21:59
    
@Sam The modulus here is the usual absolute value for real numbers. –  Austin Mohr Nov 24 '12 at 22:03
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@Sam: as I always tell my students, the square root of $(-3)^2$ is not $-3$. So $1^2<(-2)^2$ doesn't allow you to claim that $1<-2$. As for the "less than" sign, yes, that was a typo and Austin has already corrected it. –  Martin Argerami Nov 24 '12 at 22:05
    
Can that square root of 3 give a plus and a minus sign or just a positive square root? Is the RHS $|\pm \sqrt{3} \text{Re}(z)| = \sqrt{3}|\text{Re}(z)|$? –  Sam Nov 24 '12 at 22:09
    
The common practice is that when you write the square root of a positive real number, you mean the positive square root. Because of that, $\sqrt{x^2}=|x|$. –  Martin Argerami Nov 24 '12 at 22:28
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As in any other such case:

$$\forall\,x,a\in\Bbb R\,\,,\,a>0\;\;\;,\;x^2<a\Longrightarrow |x|<\sqrt a\Longleftrightarrow -\sqrt a< x<\sqrt a$$

So here

$$Im(z)^2<3\,Re(z)^2\Longrightarrow |Im(z)|<\sqrt 3\,|Re(z)|$$

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