Proof of triangular inequality for $|z_1-z_2|$ I want to prove that the following is a metric on $\Bbb C$.
$|z_1-z_2|$ where $z_1,z_2\in \Bbb C$. I have done all easily but triangular inequality:
For triangular inequality, I applied it, as if it were true, backwards on $|z_1-z_2|+|z_2-z_3|$ and got back $|z_1-z_3|\leq$, but that is a circular argument isn't it?

We can write $$|z_1-z_2|^2=(z_1-z_2)\overline{(z_1-z_2)}=(z_1-z_2)(\overline{z_1}-\overline{z_2})$$$$=z_1\overline{z_1}-z_1\overline{z_2}-z_2\overline{z_1}+z_2\overline{z_2}$$
$$ = |z_1|^2-2Re(z_1\overline{z_2})+|z_2|^2$$
$$\leq |z_1|^2-2|z_1\overline{z_2}|+|z_2|^2$$ $$=|z_1|^2-2|z_1||z_2|+|z_2|^2$$
Therefore,
$$|z_1-z_2|^2\leq(|z_1|-|z_2|)^2$$
Result follows.
