[กิตติ]

$\left|\,z_1-z_2\right|^2=(z_1-z_2)(\overline{z_1}-\overline{z_2} ) $
$=\left|\,z_1\right|^2+\left|\,z_2\right|^2-z_1\overline{z_2}-z_2 \overline{z_1} $
$=\left|\,z_1\right|^2+\left|\,z_2\right|^2-(z_1\overline{z_2}+z_2 \overline{z_1}) $

$z_1=4(\cos 145^\circ +i\sin 145^\circ ),\overline{z_1}=4(\cos 145^\circ -i\sin 145^\circ )$
$z_2=\sqrt{3} (\cos 115^\circ +i\sin 115^\circ ),\overline{z_2}=\sqrt{3} (\cos 115^\circ -i\sin 115^\circ )$

$z_1\overline{z_2}=4\sqrt{3} (\cos 145^\circ +i\sin 145^\circ )(\cos 115^\circ -i\sin 115^\circ)$
$=4\sqrt{3}((\cos 145^\circ\cos 115^\circ+\sin 145^\circ\sin 145^\circ)+i(\sin 145^\circ\cos 115^\circ-\cos 145^\circ\sin 115^\circ))$
$=4\sqrt{3}(\cos 30^\circ+i\sin 30^\circ)$

$z_2\overline{z_1}=4\sqrt{3}(\cos 115^\circ+i\sin 115^\circ) (\cos 145^\circ -i\sin 145^\circ )$
$=4\sqrt{3}((\cos 145^\circ\cos 115^\circ+\sin 145^\circ\sin 145^\circ)-i(\sin 145^\circ\cos 115^\circ-\cos 145^\circ\sin 115^\circ))$
$=4\sqrt{3}(\cos 30^\circ-i\sin 30^\circ)$

$z_1\overline{z_2}+z_2 \overline{z_1}=8\sqrt{3}\cos 30^\circ $
$=12$

$\left|\,z_1-z_2\right|^2=\left|\,z_1\right|^2+\left|\,z_2\right|^2-(z_1\overline{z_2}+z_2 \overline{z_1})$
$=16+3-12$
$=7$