Let 1, omega,and omega^2 be the cube roots of | vedclass

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

Question

(B) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$. The roots of the polynomial are given as: $z_1 =

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$. The roots of the polynomial are given as: $z_1 = 2\omega$ $z_2 = 2\omega^2$ $z_3 = 3+4\omega$ $z_4 = 3+4\omega^2$ $z_5 = 5-(\omega+\omega^2) = 5-(-1) = 6$ Since the polynomial has real coefficients,if a complex number $z$ is a root,its conjugate $\bar{z}$ must also be a root. $1$. For $z_1 = 2\omega$,the conjugate is $\bar{z_1} = 2\bar{\omega} = 2\omega^2$,which is $z_2$. $2$. For $z_3 = 3+4\omega$,the conjugate is $\bar{z_3} = 3+4\bar{\omega} = 3+4\omega^2$,which is $z_4$. $3$. For $z_5 = 6$,the conjugate is $\bar{z_5} = 6$,which is $z_5$ itself. All these roots are already present in the given set. Thus,the set of roots is $\{2\omega, 2\omega^2, 3+4\omega, 3+4\omega^2, 6\}$. There are $5$ distinct roots. Therefore,the minimum degree of the polynomial is $5$.

Let $1, \omega$,and $\omega^2$ be the cube roots of unity. The least possible degree of a polynomial with real coefficients,having $2\omega, 2\omega^2, 3+4\omega, 3+4\omega^2$,and $5-\omega-\omega^2$ as roots is

Similar Questions

One of the cube roots of unity is

$\sum_{k=1}^6 \left[ \sin \left(\frac{2 \pi k}{7}\right) - i \cos \left(\frac{2 \pi k}{7}\right) \right] = $

One of the values of $(-32 i)^{\frac{2}{5}}$ is

If $\omega$ is the complex cube root of unity and $\left(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}\right)^k+\left(\frac{a+b \omega+c \omega^2}{b+a \omega^2+c \omega}\right)^l=2$,then $2k+l$ is always

For $n \in Z^{+}$,$(1+\sin \theta+i \cos \theta)^n+(1+\sin \theta-i \cos \theta)^n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module