The number of complex roots of the equation x | vedclass

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Question

(A) The given equation is $x^{11}-x^7+x^4-1=0$. Factoring the expression: $x^7(x^4-1) + 1(x^4-1) = 0$ $(x^7+1)(x^4-1) = 0$ This implies $x^7 = -1$ or

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$2$

Vedclass · Answer

(A) The given equation is $x^{11}-x^7+x^4-1=0$. Factoring the expression: $x^7(x^4-1) + 1(x^4-1) = 0$ $(x^7+1)(x^4-1) = 0$ This implies $x^7 = -1$ or $x^4 = 1$. For $x^4 = 1$,the roots are $1, -1, i, -i$. The root $i$ has an argument of $\frac{\pi}{2}$,which is on the boundary of the first quadrant. For $x^7 = -1$,the roots are $e^{i(\frac{(2k+1)\pi}{7})}$ for $k = 0, 1, 2, 3, 4, 5, 6$. The arguments are $\frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \pi, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$. The arguments in the first quadrant $(0 Thus,there are $2$ such roots.

The number of complex roots of the equation $x^{11}-x^7+x^4-1=0$ whose arguments lie in the first quadrant is

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