$\left\vert z-2-i\right\vert$ = $\left\vert z+2-2i\right\vert$
$\left\vert (a-2)+(b+1)i\right\vert$ = $\left\vert (a+2)+(b-2)i\right\vert$
$\sqrt{(a-2)^2 + (b+1)^2}$ = $\sqrt{(a+2)^2 + (b-2)^2}$
$(a-2)^2$ + $(b+1)^2$ = $(a+2)^2$ + $(b-2)^2$
-$4a$+$2b$+$1$ = -$4a$-$4b$+$4$ 
$6b$ = $8a+3$.....$(1)$
และ $\left\vert z+1\right\vert$ = $\left\vert z+i\right\vert$
$\left\vert (a+1)+bi\right\vert$ = $\left\vert a+(b+1)i\right\vert$
$(a+1)^2$ + $(b)^2$ = $(a)^2$ + $(b+1)^2$
$2a$ = $2b$....(2)
จาก $(1)$และ$(2)$ 
แก้หาค่า $a$ = $\frac{-3}{2}$, $b$ = $\frac{-3}{2}$
จะได้ $z$ = $\frac{-3}{2}$-$\frac{3}{2}$$i$
${\left\vert 2z\right\vert}^2$ = ${\left\vert 2(\frac{-3}{2}-\frac{3}{2}i)\right\vert}^2$
${\left\vert 3-3i\right\vert}^2$
${\sqrt{(-3)^2+(-3)^2}}^2$
=${\sqrt{18}}^2$ = $18$