Question

If $z_1,z_2,z_3,z_4$ are the vertices of a square having centre at $z_0,$ then the value of ${(z_1-z_0)}^3+{(z_2-z_0)}^3+{(z_3-z_0)}^3+{(z_4-z_0)}^3$ is equal to

• A. -i
• B. i
• C. 1

Answer 1

We have $z_2-z_0=(z_1-z_0)\omega$

$z_3-z_0=(z_1-z_0){\omega }^2$
and $z_4-z_0=(z_1-z_0){\omega }^3$
Where $\omega ,{\omega }^2,{\omega }^3$are the fourth roots of unity hence,
We have
$\begin{array}{c}4\\ \Sigma \\ r=1\end{array}{(z_r-z_0)}^3={(z_1-z_0)}^3(1+{\omega }^3+{\omega }^6+{\omega }^9)$
= ${(z_1-z_0)}^3\frac{1-{\left({\omega }^3\right)}^4}{1-{\omega }^3}$
= ${(z_1-z_0)}^3\frac{1-{\left({\omega }^4\right)}^3}{1-{\omega }^3}=0$