Question

Given $z=cos\left(\frac{2\pi }{2n+1}\right)+isin\left(\frac{2\pi }{2n+1}\right)$, n a positive integer, find the equation whose roots are $\alpha =z+z^3+......+z^{2n-1}$ and $\beta =z^2+z^4+.....+z^{2n}$

• A. $z^2-z+\frac{1}{4}se{c}^2\left(\frac{\pi }{2n+1}\right)$
• B. $z^2+z+\frac{1}{4}se{c}^2\left(\frac{\pi }{2n+1}\right)$
• C. $z^2+z+\frac{1}{4}se{c}^2\left(\frac{\pi }{2n}\right)$
• D. $z^2+z+\frac{1}{2}se{c}^2\left(\frac{\pi }{2n+1}\right)$

Answer 1

The correct option is B $z^2+z+\frac{1}{4}se{c}^2\left(\frac{\pi }{2n+1}\right)$

Hence the equation will be of the type
$z^2-(\alpha +\beta )z+(\alpha .\beta )$
Now
$\alpha +\beta$
$=z+z^2+z^3...z^{2n}$
Adding and subtracting 1, we get
$\alpha +\beta$
$=[1+z+z^2+z^3...z^{2n}]-1$
$=\frac{z^{2n+1}-1}{z-1}-1$
$=-1$
Hence
$\alpha +\beta =-1$.
Therefore the equation will be of type
$z^2+z+(\alpha .\beta )$.
$\alpha =\frac{z(z^n-1)}{z-1}$ ...(i)
And
$\beta =\frac{z^2(z^n-1)}{z-1}$ ...(ii)
Hence
$\alpha .\beta =\frac{z^3.(z^n-1{)}^2}{(z-1{)}^2}$. ...(iii)
Now let us assume $n=1$.
Hence
$z=e^{i\frac{2\pi }{3}}$.
Substituting in iii we get
$\alpha .\beta =\frac{z^3.(z^n-1{)}^2}{(z-1{)}^2}$
$=z^3$ ...(n=1) in this case.
$=e^{i2\pi }$
$=1$
$=\frac{1}{4}se{c}^2\left(\frac{2\pi }{3}\right)$
Hence the equation becomes
$z^2+z+\frac{1}{4}se{c}^2\left(\frac{2\pi }{3}\right)$
Hence for $nϵN$, we get
$z^2+z+\frac{1}{4}.se{c}^2\left(\frac{2\pi }{2n+1}\right)$.