How to proof the reverse triangle inequality: $|z_{1} - z_{2}| \geq ||z_{1}|-|z_{2}||,$ I need to prove that, for any two complex numbers, $z_{1}$ and  $z_{2}$, the following is true:
$$|z_{1} - z_{2}| \geq ||z_{1}|-|z_{2}||,$$ 
Could anyone help me please?   
 A: $$|z_{1} | = |z_1-z_{2}+z_2| \leq |z_{1}-z_{2}| +|z_2| \implies |z_{1} | -|z_2| \leq |z_{1}-z_{2}|,$$
also
$$|z_{2} | = |z_2-z_{1}+z_1| \leq |z_{1}-z_{2}| +|z_1|\implies |z_{2} | -|z_1| \leq |z_{1}-z_{2}|,$$
hence 
$$ ||z_{1}|-|z_{2}|| \le |z_{1} - z_{2}|$$
A: Note that $$|z_2+(z_1-z_2)| \le |z_1-z_2|+|z_2|$$
Also, you can switch the role of $z_1$ and $z_2$.
A: with$$z_1=a+bi,z_2=c+di$$ we get
$$|a-c+i(b-d)|\geq |\sqrt{a^2+b^2}-\sqrt{c^2+d^2}|$$
and you have to prove that
$$\sqrt{(a-c)^2+(b-d)^2}\geq |\sqrt{a^2+b^2}-\sqrt{c^2+d^2}|$$
can you finish?
A: Start from the triangle inequality:-  $|z_1 + z_2| \leq |z_1| + |z_2|$.
Note that we can let $|z_1| = |(z_1 – z_2) + z_2|$. Applying the inequality to it, we have 
$|z_1| = |(z_1 – z_2) + z_2| \leq |(z_1 – z_2)| + |z_2|$.
Therefore, $|z_1| - |z_2| \leq |(z_1 – z_2)|$.
Do the same trick to $|z_2| = |(z_2 – z_1) + z_1|$ to get
$-|(z_1 – z_2)| \leq |z_1| - |z_2|$.
Combining the two results, we get
$-|(z_1 – z_2)| \leq |z_1| - |z_2| \leq |(z_1 – z_2)|$, which is the alternate form of the required.
