Question

Let $z_1,z_2,z_3$ and $z_4$ be the roots of the equation $z^7+z^4+z^3+z+1=0$ such that $z_i^3+1\ne z_i^2\forall i\in \{1,2,3,4\}$. If area of the quadrilateral formed by $z_1,z_2,z_3$ and $z_4$ on argand plane is $\left(p{\cos }^3{q}^{\circ }\right)$ where $q\in (0,90)$ then find the value of $\left(\frac{p+q}{5}\right)$.

Answer 1

Answer: 4

$z^7-z^2+z^4+z^3+z^2+z+1=0$
$\Rightarrow z^2\left(z^5-1\right)+\frac{\left(z^5-1\right)}{z-1}=0$ $\Rightarrow \left(\frac{z^3-z^2+1}{z-1}\right)\left(z^5-1\right)=0$
$\Rightarrow z^5=1$
$\text{ Area of quadrilateral }=3\times \frac{\sin 72^{\circ }}{2}+\frac{\sin 144^{\circ }}{2}$
$=\frac{1}{2}\left[3\cos 18^{\circ }+4{\cos }^{3}18^{\circ }-3\cos 18^{\circ }\right]=2{\cos }^{3}18^{\circ }$
$p=2,q=18\Rightarrow \frac{p+q}{5}=\frac{20}{5}=4$