If z1=z2=z3=1 and z1+z2+z3=0, then the area of the triangle whose vertices are z1, z2, z3 is

If $\left|{\mathrm{z}}_{1}\right|=\left|{\mathrm{z}}_{2}\right|=\left|{\mathrm{z}}_{3}\right|=1$ and ${\mathrm{z}}_{1}+{\mathrm{z}}_{2}+{\mathrm{z}}_{3}=0$, then the area of the triangle whose vertices are z1, z2, z3 is

1. A

$3\sqrt{3}/4$

2. B

$\sqrt{3}/4$

3. C

1

4. D

2

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Solution:

$\left|{\mathrm{z}}_{1}\right|=\left|{\mathrm{z}}_{2}\right|=\left|{\mathrm{z}}_{3}\right|=1$

Hence, the circumcenter of triangle is origin. Also, centroid (z1 + z2 + z3)/3 = 0, which coincides with the circumcenter. So the triangle is equilateral. Since radius is 1, length of side is $\mathrm{a}=\sqrt{3}$. Therefore, the area of the triangle is $\left(\sqrt{3}/4\right){\mathrm{a}}^{2}=\left(3\sqrt{3}/4\right)$.

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