If $\omega$ is a complex cube root of unity, then for positive integral value of $n$, the product of $\omega.\ \omega^{2}.\ \omega^{3}\ldots\ \omega^{n}$ will be

Question

Dr. Abubacker · Accepted Answer

(d)
We have, $\omega^{1} \bullet \omega^{2} \bullet \omega^{3} \bullet \ldots \bullet \omega^{n}$
$= \omega^{1 + 2 + 3 + ... + n} = \omega^{\frac{n(n + 1)}{2}} = S_{n}$ (say)
On putting $n = 1,\ 2,\ 3,\ldots,n,\$we get
$S_{1} = \omega^{1} = \omega,\ S_{2} = \omega^{3} = 1,$
$S_{3} = \omega^{6} = 1,\ldots,\ S_{7} = \omega^{28} = \omega$
$	herefore\$ We always get 1 and $\omega$

Q. If $\omega$ is a complex cube root of unity, then for positive integral value of $n$, the product of $\omega.\ \omega^{2}.\ \omega^{3}\ldots\ \omega^{n}$ will be

Explanation

Reviewed by Dr. Abubacker

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