<s>[INST] minimum value of $\bigg||z_{1}|-|z_{2}|\bigg|$ 
If $z_{1}\;,z_{2}$ are two complex number $(|z_{1}|\neq |z_{2}|)$ satisfying
$\bigg||z_{1}|-4\bigg|+\bigg||z_{2}|-4\bigg|=|z_{1}|+|z_{2}|$
$=\bigg||z_{1}|-3\bigg|+\bigg||z_{2}|-3\bigg|.$Then minimum of $\bigg||z_{1}|-|z_{2}|\bigg|$

Try: Let $|z_{1}|=a$ and $|z_{2}|=b$ and $a,b\geq 0$ and $a\neq b$
So $$|a-4|+|b-4|=|a|+|b| = |a-3|+|b-3|.$$
Let $A(0,0)$ and $B(3,3)$ and $C(4,4)$ and $P(a,b)$
Then $PA=PB=PC.$ and we have to find $|a-b|$
so i did not understand how can i conclude it,
could someone please explain me
Thanks
 [/INST] From the equality
$$|a-4|+|b-4|=a+b = |a-3|+|b-3|$$
we have that $a,b\ge 3$ is not possible.
So, assume $a\ge 3, b\le 3$. Then
$$a+b = a-3+3-b=a-b$$ which is only possible if $b=0.$ In such a case
$$|a-4|+4=a =a-3+3=a.$$ This is only possible if $a\ge 4.$
Now assume $a,b\le 3.$ In such a case
$$4-a+4-b=a+b=3-a+3-b.$$ But there is no solution.
So, $a\ge 4$ and $b=0.$
Because of the symmetry we can have $b\ge 4$ and $a=0.$
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