Basic complex number property: $z\in \mathbb{C}$, $|z|>0$ if and only if $z\neq 0$ and $|z|^2=z\bar{z}$

Question

what I've done, let $|z|=\sqrt{x^2+y^2}>0$ then we can not have x,y equally zero hence $z\neq 0$ , now $z\bar{z}=x^2+y^2=|z|^2$.

conversely, let $z\neq 0$ and $|z|^2=z\bar{z}$, then $x=0,$or $y=0$ but not the at the same time,hence $|z|>0$.

Answer 1

For every $z\in\Bbb C$, you have $|z|^2=z\overline z$ since, if $z=a+bi$ ($a,b\in\Bbb R)$, then\begin{align}|z|^2&=a^2+b^2\\&=(a+bi)(a-bi)\\&=z\overline z.\end{align}So, the problem becomes this: to prove that $|z|>0\iff z\ne0$. If $z=0$, then neither $|z|>0$ nor $z\ne0$ hold. And if $z\ne0$, then both assertions hold. Therefore, they are equivalent.