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Home » e 1 $, and $ \omega^2 + \omega + 1 = 0 $. - AMAZONAWS

e 1 $, and $ \omega^2 + \omega + 1 = 0 $. - AMAZONAWS

📅 March 6, 2026 👤 scraface
Mar 06, 2026
e 1 $, and $ \omega^2 + \omega + 1 = 0 $.

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  • So $ f(-y) = f(y) $, hence $ f $ is even.

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