Question

If $\omega (\ne 1)$ is a cube root of unity and ${(1+\omega )}^7=A+B\omega$. Then $(A,B)$ equals to

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

Answer 1

The correct option is A $(1,1)$

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

$1+\omega +{\omega }^2=0\therefore 1+\omega =-{\omega }^2,{\omega }^3=1$

$(-{\omega }^2{)}^7=A+B\omega$

$-{\omega }^{12}.{\omega }^2=A+B\omega .({\omega }^{12}=({\omega }^3{)}^4=1^4=1)$

$-{\omega }^2=A+B\omega .$

$1+\omega =A+B\omega$ .(comparing)

$\therefore$ A=1 B=1