Question

If $\alpha =\cos \left(\frac{\pi }{n}\right)+i\sin \left(\frac{\pi }{n}\right)$, then
$(x-1)(x+1)(x-\alpha )(x-\overline{\alpha })...(x-{\alpha }^{n-1})(x-{\overline{\alpha }}^{n-1})$ equals

• A. $x^n-1$
• B. $x^n+1$
• C. $x^{2n}+1$
• D. $x^{2n}-1$

Answer 1

The correct option is D $x^{2n}-1$

$f\left(x\right)=a_0{x}^m+a_1{x}^{m-1}+.....+a_{m-1}x+a_m=0$

then, $f\left(x\right)=a_0(x-{\alpha }_1)......(x-{\alpha }_m)$

and $\frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}=\frac{1}{x-{\alpha }_1}+....+\frac{1}{x-{\alpha }_m}$

$\Rightarrow \alpha =\cos \left(\frac{\pi }{x}\right)+i\sin \left(\frac{\pi }{x}\right)$ then,

$\Rightarrow (x-1)(x+1)(x-\alpha )(x-\overline{\alpha }).....(x-{\alpha }^{n-1})(x-{\alpha }^{-n-1})$

$=x^{(n-1+n+1)}-1$

$=x^{\left(2n\right)}-1.$

Hence, the answer is $x^{\left(2n\right)}-1.$