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

The minimum for $|a-b|$ is at $a=b$
$$
|a-c|-a = |b-c|-b \Rightarrow a+b=c
$$
so the intersections for
$$
a+b = 4 \cap a = b \Rightarrow a = 2\\
a+b = 3 \cap a = b \Rightarrow a = \frac 32
$$
so the minimum is at $a = b = \frac 32$ or $a = b = 2$ which is $0$
Attached in red $a+b=4$ in blue $a+b=3$ and in lightblue $a = b$

