The number of complex roots of the equation x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

$2$

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

Similar Questions

If $\omega$ is a complex cube root of unity,then $\left(1-\omega+\omega^2\right)^6+\left(1-\omega^2+\omega\right)^6=$

The value of $\left[ \frac{1 - \cos \frac{\pi}{10} + i\sin \frac{\pi}{10}}{1 - \cos \frac{\pi}{10} - i\sin \frac{\pi}{10}} \right]^{10} = $

If the roots of the equation $(z-4)^3=8 i$ are $a-2 i, b+i$,and $c+i$,then $\sqrt{a b c}=$

The value of $\sum\limits_{r = 1}^8 {\left( {\sin \frac{{2r\pi }}{9} + i\cos \frac{{2r\pi }}{9}} \right)} $ is

If $z^{2} + z + 1 = 0$,$z \in \mathbb{C}$,then $\left| \sum_{n=1}^{15} \left( z^{n} + (-1)^{n} \frac{1}{z^{n}} \right)^{2} \right|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module