TITLE: How to prove $|z_1-z_2| \geq |z_1|-|z_2|$ in other way than this?
QUESTION [8 upvotes]: How to prove $|z_1-z_2| \geq |z_1|-|z_2|$ in other way than this? I mean I tried to find on the internet but could not find. I ask for more straighforward way than the proof that is presented for item 3.

REPLY [19 votes]: Use the triangle inequality on $|z_1| = |(z_1 - z_2) + (z_2)|$.

REPLY [2 votes]: Let $z_1=r_1(\cos A+i\sin A)$  and $z_2=r_2(\cos B+i\sin B)$
So, $|z_1|=r_1$ and $|z_2|=r_2$
$|z_1-z_2|$
$=\sqrt{(r_1\cos A-r_2\cos B)^2+(r_1\sin A-r_2\sin B)^2}$
$=\sqrt{r_1^2+r_2^2-2r_1r_2\cos(A-B)}$
$≥\sqrt{r_1^2+r_2^2-2r_1r_2}$ as $\cos(A-B)≤1$
$=r_1-r_2$
So, $|z_1-z_2|≥|z_1|-|z_2|$, the equality occurs when  $\cos(A-B)=1$ i.e., when $A=B$