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Question Number 143970 by ZiYangLee last updated on 20/Jun/21

$$\mathrm{Given}\mathrm{that}\omega \mathrm{is}\mathrm{a}\mathrm{complex}\mathrm{number},$$$${\omega }^7=1,\omega \ne 1,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$${\omega }^1+{\omega }^2+{\omega }^3+{\omega }^4+{\omega }^5+{\omega }^6.$$

Answered by Rasheed.Sindhi last updated on 20/Jun/21

$$Sumofallthenthroots=0$$$$1+{\omega }^1+{\omega }^2+{\omega }^3+{\omega }^4+{\omega }^5+{\omega }^6=0$$$${\omega }^1+{\omega }^2+{\omega }^3+{\omega }^4+{\omega }^5+{\omega }^6=-1$$

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