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Question Number 143970 by ZiYangLee last updated on 20/Jun/21
Given that ω is a complex number,  ω^7 =1, ω≠1, find the value of  ω^1 +ω^2 +ω^3 +ω^4 +ω^5 +ω^6 .
$$\mathrm{Given}\:\mathrm{that}\:\omega\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}, \\ $$$$\omega^{\mathrm{7}} =\mathrm{1},\:\omega\neq\mathrm{1},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\omega^{\mathrm{1}} +\omega^{\mathrm{2}} +\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{5}} +\omega^{\mathrm{6}} . \\ $$
Answered by Rasheed.Sindhi last updated on 20/Jun/21
Sum of all the nth roots=0  1+ω^1 +ω^2 +ω^3 +ω^4 +ω^5 +ω^6 =0  ω^1 +ω^2 +ω^3 +ω^4 +ω^5 +ω^6 =−1
$${Sum}\:{of}\:{all}\:{the}\:{nth}\:{roots}=\mathrm{0} \\ $$$$\mathrm{1}+\omega^{\mathrm{1}} +\omega^{\mathrm{2}} +\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{5}} +\omega^{\mathrm{6}} =\mathrm{0} \\ $$$$\omega^{\mathrm{1}} +\omega^{\mathrm{2}} +\omega^{\mathrm{3}} +\omega^{\mathrm{4}} +\omega^{\mathrm{5}} +\omega^{\mathrm{6}} =−\mathrm{1} \\ $$

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