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I really couldn't understand last part when they pass from $|z|^{2}$ to $|z|$ so any more explanation please?

complex numbers

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Perhaps $$z\overline z=|z|^2\Longleftrightarrow \frac{|z|^2}{z}=\overline z\,\,,\,\,z\neq 0\,\,\,?$$ –  DonAntonio Jan 23 '13 at 15:13
    
I'm looking for $|z|$ not $\overline z$ nor $z$ –  pourjour Jan 23 '13 at 15:18
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1 Answer 1

up vote 3 down vote accepted

Simply expand: $$\left(-a + \sqrt{a^2+4}\right)^2 = 2\left(a^2 + 2 - \sqrt{a^4+4a^2}\right)$$ If $|z|^2 > r^2$ it follows that $|z|>r$...

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I didn't get the point of expanding this expression. –  pourjour Jan 23 '13 at 15:21
    
@pourjour - the two expression are exactly the ones you need, are they not? –  nbubis Jan 23 '13 at 15:22
    
ok I get it thanks –  pourjour Jan 23 '13 at 15:25
    
Maybe clearer to write: $$\left(\frac{-a + \sqrt{a^2+4}}2\right)^2 = \left(\frac{a^2 + 2 - \sqrt{a^4+4a^2}}{2}\right)$$ –  Thomas Andrews Jan 23 '13 at 15:26
    
how about doing this by respecting the steps and not reversly? –  pourjour Jan 23 '13 at 15:26
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