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Teaching myself Linear Algebra and got stuck on the following question for complex numbers:
Show that $|z| \geq 0$ and $|z|=0$ if and only if $z=0.$
Now, the question itself seems pretty obvious where $z=x+iy$ and $|z|=\sqrt{x^2+y^2}$ but I am a bit confused regarding how to tie everything together. Thanks in advance.

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    $\begingroup$ Write $|z|^2=z\,\bar z$. $\endgroup$
    – lhf
    May 30, 2013 at 1:32
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    $\begingroup$ You can use \geq for getting $\geq$ $\endgroup$
    – Belgi
    May 30, 2013 at 1:33

1 Answer 1

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Now, $|Z|= 0$ implies that $\sqrt{x^2+y^2}=0$ which implies that $x^2+y^2=0$. Under what circumstances will the last equality hold?

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    $\begingroup$ Both $x$ and $y$ have to be real btw. $\endgroup$
    – Shuhao Cao
    May 30, 2013 at 2:16

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