Question

If $\omega$ $(\ne 1)$ is a cube root of unity and ${(1+\omega )}^7=A+B\omega$ then $A$ and $B$ are respectively.

• A. $0,1$
• B. $1,1$
• C. $1,0$
• D. $-1,1$

Answer 1

The correct option is B

$1,1$

Apply the properties of the cube root of unity of complex numbers.

$\therefore {\omega }^3=1\text{and}1+\omega +{\omega }^2=0$

Now, ${(1+\omega )}^7=A+B\omega$

$\begin{array}{l}\Rightarrow {(-{\omega }^2)}^7=A+B\omega \left[\because 1+\omega =-{\omega }^2\right]\\ \Rightarrow (-{\omega }^{14})=A+B\omega \left[\because {\left(a^n\right)}^m=a^{n\times m}\right]\\ \Rightarrow -({\omega }^3\times {\omega }^3\times {\omega }^3\times {\omega }^3\times {\omega }^2)=A+B\omega [\because {\omega }^3=1]\\ \Rightarrow -\left({\omega }^2\right)=A+B\omega \\ \Rightarrow 1+\omega =A+B\omega \end{array}$

Now, compare the left-hand side to the right-hand side, and we get

$A=1\mathrm{and}B=1$

Hence, the correct answer is an option (B).