Mathematics
curtismanjones
2017-08-12 23:14:45
Prove that w^49+w^101+w^150=0,where w is a complex cube root of unity
ANSWERS
Lukedunn123
2017-08-13 05:17:10

Because [latex]w[/latex] is a cube root of unity, you have [latex]w^3=1[/latex]. So [latex]w^{49}=w(w^3)^{16}=w[/latex] [latex]w^{101}=w^2(w^3)^{33}=w^2[/latex] [latex]w^{150}=(w^3)^{50}=1[/latex] and so [latex]w^{49}+w^{101}+w^{150}=1+w+w^2=dfrac{1-w^3}{1-w}[/latex] provided that [latex]w eq1[/latex]. Any other cube root of unity will make the numerator vanish, so the equality holds.

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