[กิตติ]

อีกวิธีหนึ่ง
$\left|\,z_1-z_2\right|=\left|\,z_1\right| \left|\,1-\frac{z_2}{z_1} \right| $
$\frac{z_2}{z_1}=\frac{\sqrt{3} (\cos 115^\circ +i\sin 115^\circ )}{4(\cos 145^\circ +i\sin 145^\circ )} $
$=\frac{\sqrt{3}}{4} (\cos (-30^\circ) +i\sin(-30^\circ))$
$=\frac{\sqrt{3}}{4} (\cos (30^\circ) -i\sin(30^\circ))$
$=\frac{3}{8}-i\frac{\sqrt{3}}{8} $

$1-\frac{z_2}{z_1}=1-(\frac{3}{8}-i\frac{\sqrt{3}}{8})=\frac{5}{8}+i\frac{\sqrt{3}}{8} $
$\left|\,1-\frac{z_2}{z_1}\right|^2=\frac{7}{16} $

$\left|\,z_1-z_2\right|^2=\left|\,z_1\right|^2 \left|\,1-\frac{z_2}{z_1} \right|^2 $
$\left|\,z_1\right|^2 =16$
$\left|\,z_1-z_2\right|^2=7$