Let z₁, z₂, z₃ be the complex numbers representing the vertices A, B, C of a triangle described in counterclock sense respectively. Consider the following statements.
$\text{ I) }\mathrm{△}\mathrm{ABC}$ is equilateral               $\text{ II) }{\mathrm{z}}_3-{\mathrm{z}}_1=\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)\left(\cos \frac{\mathrm{\pi }}{3}+\mathrm{isin}\frac{\mathrm{\pi }}{3}\right)$
$\text{ III) }{\mathrm{z}}_2-{\mathrm{z}}_1=\left({\mathrm{z}}_3-{\mathrm{z}}_1\right)\mathrm{cis}\frac{5\mathrm{\pi }}{3}$     $\text{ IV) }{\mathrm{z}}_1+{\mathrm{z}}_2\mathrm{cis}\frac{2\mathrm{\pi }}{3}+{\mathrm{z}}_3\mathrm{cis}\frac{4\mathrm{\pi }}{3}=0$
Then which one is correct:

I) Suppose $∆$ABC is equilateral
$\frac{{\mathrm{z}}_3-{\mathrm{z}}_1}{{\mathrm{z}}_2-{\mathrm{z}}_1}=\cos \frac{\mathrm{\pi }}{3}+\mathrm{isin}\frac{\mathrm{\pi }}{3}$
Therefore, I$\Rightarrow$ II. (a) is true.
II) Assume that ${\mathrm{z}}_3-{\mathrm{z}}_1=\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)\left(\cos \frac{\mathrm{\pi }}{3}+\mathrm{isin}\frac{\mathrm{\pi }}{3}\right)$
Therefore $\frac{{\mathrm{z}}_3-{\mathrm{z}}_1}{{\mathrm{z}}_2-{\mathrm{z}}_1}=\cos \frac{\mathrm{\pi }}{3}+\mathrm{isin}\frac{\mathrm{\pi }}{3}$
$\Rightarrow \left|{\mathrm{z}}_3-{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2-{\mathrm{z}}_1\right|\text{ and }\mathrm{\angle }\mathrm{BAC}=\frac{\mathrm{\pi }}{3}$
This implies $∆$ABC is equilateral.
Therefore, II $\Rightarrow$I. Now rotate $\stackrel{\mathrm{\_}}{\mathrm{AC}}$ about A through angle 5$\mathrm{\pi }$ / 3 in anticlock sense so that
${\mathrm{z}}_2-{\mathrm{z}}_1=\left({\mathrm{z}}_3-{\mathrm{z}}_1\right)\left(\cos \frac{5\mathrm{\pi }}{3}+\mathrm{isin}\frac{5\mathrm{\pi }}{3}\right).$
This means II$\Rightarrow$ III.
Similarly we can see that III$\Rightarrow$IV and IV$\Rightarrow$ I.