# How to show that $|z| \geq 0$ and $|z|=0$ if and only if $z=0$?

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.

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

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?
• Both $x$ and $y$ have to be real btw. May 30, 2013 at 2:16