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 $z _{ i }^3+1 \neq z _{ i }^2 \forall i \in\{1,2,3,4\}$. 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

$ 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} $
$=\frac{1}{2}\left[3 \cos 18^{\circ}+4 \cos ^3 18^{\circ}-3 \cos 18^{\circ}\right]=2 \cos ^3 18^{\circ}$
$p =2, q =18 \Rightarrow \frac{ p + q }{5}=\frac{20}{5}=4$