If $z_1+ z_2 + z_3 = 0$ and $|z_1|=|z_2|=|z_3|= 1$, then area of triangle whose vertices are $z_1, z_2$ and $z_3$ is:

${|z^{}_{1}+z^{}_{2}|}^2+{|z^{}_{1}-z^{}_{2}|}^2=2({|z^{}_{1}|}^2+{|z^{}_{2}|}^2)$ ....... $(1)$

$z^{}_{1}+z^{}_{2}+z^{}_{3}=0$

$z^{}_{3}=-z^{}_{2}-z^{}_{1}$

By taking modulus both sides

$|-z^{}_{3}|=|z^{}_{2}+z^{}_{1}|$ .........

$(2)$

Using $(2)$ in $(1),$ we have

${|-z_{3}|}^2+{|z^{}_{1}-z^{}_{2}|^2}=2{(1+1)}$

$1+{|z^{}_{1}-z^{}_{2}|^2}=2{(1+1)}$

${|z^{}_{1}-z^{}_{2}|^2}=3$

$|z^{}_{1}-z^{}_{2}|=\sqrt{3}$

Similarily

$|z^{}_{2}-z^{}_{3}|$=

$|z^{}_{3}-z^{}_{1}|=\sqrt{3}$

Hence sides of triangle formed are $\sqrt{3},$

$\sqrt{3},$

$\sqrt{3}$ and are equal.

So triangle formed is an equilateral triangle

Area of an

equilateral triangle $=\dfrac{\sqrt{3}}{4}s^2$

where $s$ is side of

equilateral triangle

So area of triangle formed by

$z^{}_{1} ,z^{}_{2},z^{}_{3}=3\dfrac{\sqrt{3}}{4}$