complex number passing from $|z|^{2}$ to $|z|$

Question

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

complex numbers

Perhaps $$z\overline z=|z|^2\Longleftrightarrow \frac{|z|^2}{z}=\overline z\,\,,\,\,z\neq 0\,\,\,?$$

nbubis · Accepted Answer · 2013-01-23 15:18:39Z

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 $|z|>r$...

answered Jan 23 at 15:18

nbubis
6,6592934

	I didn't get the point of expanding this expression. – pourjour Jan 23 at 15:21
	@pourjour - the two expression are exactly the ones you need, are they not? – nbubis Jan 23 at 15:22
	ok I get it thanks – pourjour Jan 23 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 at 15:26
	how about doing this by respecting the steps and not reversly? – pourjour Jan 23 at 15:26

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