Complex numbers/Polynomials? (1 Viewer)

[ashokkumar]

Let w = cos(2pi/7) + isin(2pi/7). Prove that 1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0

 

[Rabbi_Nigger]

Just simply substitute, w into the equation, using de moivre's theorem, youll get a sweet nigga ass simplification, do simple addition, and subtraction, lhs = 0

 

[Trebla]

Let w = cos(2pi/7) + isin(2pi/7). Prove that 1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0

Using DeMoivre's theorem we have: w⁷ = 1
1 + w + w² + w³ + w⁴ + w⁵ + w⁶
= (w⁷ - 1)/(w - 1)
as it is a geometric series
but since w⁷ = 1 then it becomes zero

 

[Rabbi_Nigger]

Using DeMoivre's theorem we have: w⁷ = 1
1 + w + w² + w³ + w⁴ + w⁵ + w⁶
= (w⁷ - 1)/(w - 1)
as it is a geometric series
but since w⁷ = 1 then it becomes zero

OI!!!! What about my way!!!?

 

[ashokkumar]

thanks again Trebla

 

[nikkifc]

You could also use sum of roots = 0 by consider w⁷=1 as a polynomial equation and by noting that there 7 distinct roots are 1, w, w², ... , w⁷.

 

[gurmies]

You could also use sum of roots = 0 by consider w⁷=1 as a polynomial equation and by noting that there 7 distinct roots are 1, w, w², ... , w⁷.

8 roots there

 

[nikkifc]

8 roots there

Yeah my bad.