Question

One of the complex roots of the equation $x^{11}-x^{6}-x^{5}+1=0$ is

• A. cis $ \frac{3 \pi}{5}$
• B. cis $\frac{\pi}{3}$
• C. cis $\frac{5 \pi}{6}$
• D. cis $\frac{7 \pi}{5}$

Answer 1

Answer: (B)

$x^{11}-x^{6}-x^{5}+1=0$
$\Rightarrow \,\left(x^{6}-1\right)\left(x^{5}-1\right)=0$
$\Rightarrow \, x^{6}=1 \text { or } x^{5}=1$
$\Rightarrow \, x=(1)^{\frac{1}{6}} \text { or } x=(1)^{\frac{1}{5}}$
$x=\cos \frac{2 k \pi}{6}+i \sin \frac{2 k \pi}{6} \text { or } x=\cos \frac{2 r \pi}{5}+i \,\sin \frac{2 r \pi}{5}$
where, $k=0,1,2,3,4,5$ where $r=0,1,2,3,4$
$x=\cos \frac{k \pi}{3}+i \sin \frac{k \pi}{3}$
When $k=1$
$\cos \frac{\pi}{3}+i \sin \pi / 3=C \text { is } \frac{\pi}{3}$