If ${\mathrm{Z}}_1,{\mathrm{Z}}_2,{\mathrm{Z}}_3,{\mathrm{Z}}_4$ are the roots of the equation ${\mathrm{z}}^4+{\mathrm{z}}^3+{\mathrm{z}}^2+\mathrm{z}+1=0$, then
 where [ ] represents greatest integer function

The given equation is $\frac{{\mathrm{z}}^5+1}{\mathrm{z}-1}=0$which means that ${\text{z}}_1,$ ${\mathrm{z}}_2,{\mathrm{z}}_3,{\mathrm{z}}_4$ are four out of five roots of unity except 1.
$\text{ A) }{\mathrm{z}}_1^4+{\mathrm{z}}_2^4+{\mathrm{z}}_3^4+{\mathrm{z}}_4^4+1^4=0\Rightarrow \left|\sum _{\mathrm{i}=1}^4 {\mathrm{z}}_1^4\right|=1$
$\text{ B) }{\mathrm{z}}_1^5+{\mathrm{z}}_2^5+{\mathrm{z}}_3^5+{\mathrm{z}}_4^5+1^5=5\Rightarrow \left|\sum _{\mathrm{i}=1}^4 {\mathrm{z}}_1^5\right|=4$
$\text{ C) }{\mathrm{z}}^4+{\mathrm{z}}^3+{\mathrm{z}}^2+\mathrm{z}+1=\left(\mathrm{z}-{\mathrm{z}}_1\right)\left(\mathrm{z}-{\mathrm{z}}_2\right)\left(\mathrm{z}-{\mathrm{z}}_3\right)\left(\mathrm{z}-{\mathrm{z}}_4\right)$
Putting z = –2 both sides and we get $\prod _{\mathrm{i}=1}^4 \left({\mathrm{z}}_1+2\right)=11$
$\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|=\sqrt{2+2\cos {144}^0}$ for minimum 
$=2\cos {72}^0=\frac{\sqrt{5}-1}{2}$ whose greatest integer is 0