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)

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