Question

If $z_1,z_2,z_3$ and $z_4$ are the roots of the equation $z^4+z^3+z^2+z+1=0$, then


$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(A)}& \left|\sum _{i=1}^4(z_i{)}^4\right|\text{is equal to}& \text{(p)}& 0\\ \text{(B)}& \sum _{i=1}^4(z_i{)}^5\text{is equal to}& \text{(q)}& 4\\ \text{(C)}& \prod _{i=1}^4(z_i+2)\text{is equal to}& \text{(r)}& 1\\ \text{(D)}& \text{Least value of}[|z_1+z_2|]\text{is}& \text{(s)}& 11\end{array}$

$\text{where}\left[\right]$ represents greatest integer function

Which of the following is the correct combination

• A. $\left(A\right)\to \left(p\right),\left(B\right)\to \left(r\right),\left(C\right)\to \left(s\right),\left(D\right)\to \left(q\right)$
• B. $\left(A\right)\to \left(r\right),\left(B\right)\to \left(q\right),\left(C\right)\to \left(s\right),\left(D\right)\to \left(p\right)$
• C. $\left(A\right)\to \left(q\right),\left(B\right)\to \left(r\right),\left(C\right)\to \left(s\right),\left(D\right)\to \left(p\right)$
• D. $\left(A\right)\to \left(p\right),\left(B\right)\to \left(q\right),\left(C\right)\to \left(r\right),\left(D\right)\to \left(s\right)$

Answer 1

The correct option is B $\left(A\right)\to \left(r\right),\left(B\right)\to \left(q\right),\left(C\right)\to \left(s\right),\left(D\right)\to \left(p\right)$

The given equation is $z^4+z^3+z^2+z+1=0$

So, the equation is $\frac{z^5-1}{z-1}$ which means that $z_1,z_2,z_3,z_4$ are four out of five roots of unity except 1

$\left(A\right)z_1^4+z_2^4+z_3^4+z_4^4+1^4=0$

$\Rightarrow \left|\sum _{i=1}^4(z_i{)}^4\right|=1$

$\left(B\right)z_1^5+z_2^5+z_3^5+z_4^5+1=5$

$\Rightarrow \sum _{i=1}^4(z_i{)}^5=4$

$\left(C\right)z^4+z^3+z^2+z+1=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$

Putting $z=-2$ on both the sides, we get

$\Rightarrow \prod _{i=1}^4(z_i+2)=11$


$\left(D\right)|z_1+z_2|=\sqrt{|z_1{|}^2+|z_2{|}^2+2|z_1||z_2|\cos ({\theta }_1-{\theta }_2)}\Rightarrow |z_1+z_2|=\sqrt{2+2\cos ({\theta }_1-{\theta }_2)}$
$\Rightarrow |z_1+z_2|=2\left|\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)\right|\Rightarrow |z_1+z_2|=2\left|\cos \left(\frac{2\pi (k_1-k_2)}{2n}\right)\right|,k_1,k_2=1,2,3,4,[k_1\ne k_2]$
Now $n=5$ and for least value $k_1-k_2=3$
$\therefore |z_1+z_2|=2\left|\cos \frac{3\pi }{5}\right|=2\cos \frac{2\pi }{5}=\frac{\sqrt{5}-1}{2}$
whose greatest integer is $0$