Q49 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics
If $1, \omega, \omega^2$ are the cube roots of unity and $1, \alpha, \alpha^2, \alpha^3$ are the fourth roots of unity in usual notation then $\alpha+\alpha \omega-\alpha^3 \omega^2=$
- A3
- B1
- Correct answerC0
- D-1
Solution
Given $1, \omega, \omega^2$ are cube roots of unity and $1, \alpha, \alpha^2$, $\alpha^3$ are fourth roots of varity.
We know that four roots of unity are $1, i,-1,-\mathrm{i}$ Now, $\alpha+\alpha \omega-\alpha^3 \omega^2=\mathrm{i}(1+\omega)+i \omega^2$ $=\mathrm{i}\left(1+\omega+\omega^2\right)$
Here, $1+\omega+\omega^2=0$
$$
\alpha+\alpha \omega-\alpha^3 \omega^2=\text { i. }(0)=0
$$
Hence, the correct answer is option: 'C'.