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One of the complex roots of the equation $x^{11}-x^6-x^5+1=0$ is
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Verified Answer
The correct answer is:
$\operatorname{cis} \frac{\pi}{3}$
$\begin{aligned} & \quad x^{11}-x^6-x^5+1=0 \\ & \Rightarrow \quad\left(x^6-1\right)\left(x^5-1\right)=0 \\ & \Rightarrow \quad x^6=1 \text { or } x^5=1 \\ & \Rightarrow \quad x=(1)^{\frac{1}{6}} \text { or } x=(1)^{\frac{1}{5}} \\ & x=\cos \frac{2 k \pi}{6}+i \sin \frac{2 k \pi}{6} \text { or } x=\cos \frac{2 r \pi}{5}+i \sin \frac{2 r \pi}{5} \\ & \end{aligned}$
where,
$$
\begin{aligned}
k & =0,1,2,3,4,5 \text { where } r=0,1,2,3,4 \\
x & =\cos \frac{k \pi}{3}+i \sin \frac{k \pi}{3}
\end{aligned}
$$
When $k=1$,
$$
\cos \frac{\pi}{3}+i \sin \pi / 3=C \text { is } \frac{\pi}{3} \text {. }
$$
where,
$$
\begin{aligned}
k & =0,1,2,3,4,5 \text { where } r=0,1,2,3,4 \\
x & =\cos \frac{k \pi}{3}+i \sin \frac{k \pi}{3}
\end{aligned}
$$
When $k=1$,
$$
\cos \frac{\pi}{3}+i \sin \pi / 3=C \text { is } \frac{\pi}{3} \text {. }
$$
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