How to prove $|z_1-z_2| \geq |z_1|-|z_2|$ in other way than this?
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.
$\endgroup$ 33 Answers
$\begingroup$Use the triangle inequality on $|z_1| = |(z_1 - z_2) + (z_2)|$.
$\endgroup$ 4 $\begingroup$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 $A=B$
$\endgroup$ $\begingroup$You can use multiplication by conjugate.
$|Z_1 + Z_2 |^2 = (Z_1 + Z_2)\overline{(Z_1 + Z_2)}$. By expansion you will get
$| Z_1 + Z_2 |^2 = |Z_1|^2 + |Z_2|^2$ + 2 Real part of $(Z_1 \overline{Z_2})$ $\leq (|Z_1| + |Z_2|)^2$.
Then use this result, and prove remaining.
$\endgroup$ 6
