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

Let $i=\sqrt{-1}$. If $\frac{(-1+i \sqrt{3})^{21}}{(1-i)^{24}}+\frac{(1+i \sqrt{3})^{21}}{(1+i)^{24}}=k$,and $n =[| k |]$ be the greatest integral part of $| k |$. Then $\sum_{ j =0}^{ n +5}( j +5)^{2}-\sum_{ j =0}^{ n +5}( j +5)$ is equal to ........ .

If $\alpha, \beta$ are non-real cube roots of $2$,then $\alpha^6 + \beta^6$ equals

The value of $\sum\limits_{k = 1}^6 {\left( {\sin \frac{{2\pi k}}{7} - i\cos \frac{{2\pi k}}{7}} \right)} $ is

If $x = \sqrt{-1 - \sqrt{-1 - \sqrt{-1 - ... \infty}}}$, where $\omega$ is a non-real complex cube root of unity, then the value of $x$ is...

$\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module