minimum value of $\bigg||z_{1}|-|z_{2}|\bigg|$

Question

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

Answer 1

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 $a=0.$

Answer 2

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 $a = b$