One of the complex roots of the equation x^{1 | vedclass

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(B) Given the equation $x^{11}-x^6-x^5+1=0$. Factorizing the expression: $x^6(x^5-1) - 1(x^5-1) = 0$ $(x^6-1)(x^5-1) = 0$ This implies $x^6=1$ or $x^5

Vedclass · Accepted Answer

$\operatorname{cis} \frac{\pi}{3}$

Vedclass · Answer

(B) Given the equation $x^{11}-x^6-x^5+1=0$. Factorizing the expression: $x^6(x^5-1) - 1(x^5-1) = 0$ $(x^6-1)(x^5-1) = 0$ This implies $x^6=1$ or $x^5=1$. The roots are given by $x = \operatorname{cis}(\frac{2k\pi}{6})$ for $k \in \{0, 1, 2, 3, 4, 5\}$ or $x = \operatorname{cis}(\frac{2r\pi}{5})$ for $r \in \{0, 1, 2, 3, 4\}$. For $k=1$,$x = \operatorname{cis}(\frac{2\pi}{6}) = \operatorname{cis}(\frac{\pi}{3})$. Thus,$\operatorname{cis}(\frac{\pi}{3})$ is one of the complex roots.

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

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