Question

The roots of equation $a{x}^2+bx+c=0$ are $\alpha =z^2+z^4+.....+z^{2n}$ and $\beta =z+z^3+.....+z^{2n-1}$, where $z=e^{i\left(\frac{2\pi }{2n+1}\right)},n\in N,$ then

• A. $\alpha +\beta =-1$
• B. $\alpha +\beta =1$
• C. Given equation can be written as $x^2+x+\frac{1}{4}{\sec }^2\left(\frac{\pi }{2n+1}\right)=0$
• D. Given equation can be written as $x^2+x+\frac{1}{4}{\cos }^2\left(\frac{\pi }{2n+1}\right)=0$

Answer 1

Answer: (A), (C)

The correct options are
A $\alpha +\beta =-1$
C Given equation can be written as $x^2+x+\frac{1}{4}{\sec }^2\left(\frac{\pi }{2n+1}\right)=0$

$z=e^{i\left(\frac{2\pi }{2n+1}\right)}$
$1,z,z^2,...,z^{2n}$ denotes the $(2n+1{)}^{\text{th}}$ roots of unity

Sum of roots of the given equation
$\alpha +\beta =z+z^2+z^3+....+z^{2n}=-1$

Product of roots
$\alpha \beta =(z^2+z^4+....z^{2n})(z+z^3+....+z^{2n-1})=z^3(1+z^2+z^4+.....+z^{2n-2}{)}^2=z^3\cdot {\left(\frac{z^{2n}-1}{z^2-1}\right)}^2=z^3\cdot {\left(\frac{z^{2n}-1}{z-1}\right)}^2\cdot \frac{1}{(z+1{)}^2}$

Now,
$1+z+z^2+z^3+....+z^{2n}=\frac{z^{2n+1}-1}{z-1}=0\Rightarrow \frac{z^{2n+1}-z+z-1}{z-1}=0\Rightarrow \frac{z(z^{2n}-1)}{z-1}=-1\Rightarrow \frac{z^{2n}-1}{z-1}=-\frac{1}{z}$

Now,
$\alpha \beta =\frac{z}{(z+1{)}^2}=\frac{1}{z+\frac{1}{z}+2}=\frac{1}{2\cos \theta +2},\text{where}\theta =\frac{2\pi }{2n+1}=\frac{{\sec }^2\left(\frac{\theta }{2}\right)}{4}$

Equation can be wriiten as $x^2-(\alpha +\beta )x=\alpha \beta =0\Rightarrow x^2+x+\frac{1}{4}{\sec }^2\left(\frac{\pi }{2n+1}\right)=0$