The sum of the complex roots of the equation | vedclass

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

Question

(A) Given equation: $(x-1)^3+64=0$ $\Rightarrow (x-1)^3 = -64$ $\Rightarrow (x-1)^3 = (-4)^3$ Let $y = x-1$,then $y^3 = (-4)^3$. The roots are $y = -4

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Given equation: $(x-1)^3+64=0$ $\Rightarrow (x-1)^3 = -64$ $\Rightarrow (x-1)^3 = (-4)^3$ Let $y = x-1$,then $y^3 = (-4)^3$. The roots are $y = -4, -4\omega, -4\omega^2$,where $\omega$ is the complex cube root of unity. Thus,$x-1 = -4, -4\omega, -4\omega^2$. The roots are $x_1 = -3$,$x_2 = 1-4\omega$,and $x_3 = 1-4\omega^2$. The complex roots are $x_2 = 1-4\omega$ and $x_3 = 1-4\omega^2$. Sum of complex roots $= (1-4\omega) + (1-4\omega^2) = 2 - 4(\omega + \omega^2)$. Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$. Sum $= 2 - 4(-1) = 2 + 4 = 6$.

The sum of the complex roots of the equation $(x-1)^3+64=0$ is

Similar Questions

If $z$ is a complex number such that $z^2+z+1=0$,then $\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3+\left(z^3+\frac{1}{z^3}\right)^3+\ldots+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3=$

If $x=a+b$,$y=a \alpha+b \beta$,$z=a \beta+b \alpha$ and $\alpha, \beta$ are the complex cube roots of unity,then $x^3+y^3+z^3=$

The $n^{th}$ roots of unity are in

If $\theta = \frac{\pi}{6}$,then the $10^{th}$ term of the series $1 + (\cos \theta + i \sin \theta) + (\cos \theta + i \sin \theta)^2 + (\cos \theta + i \sin \theta)^3 + \ldots$ is equal to:

If $z_1$ and $z_2$ are two of the $n^{\text{th}}$ roots of unity such that the line segment joining them subtends a right angle at the origin,then for a positive integer $k$,$n$ takes the form

Vedclass Test Series

Exam Paper Generator

Online Exam Module