An algebra problem by Soumya Shrivastva

Algebra Level 2

If ω \omega is the fifth root of unity, then log 2 1 + ω + ω 2 + ω 3 ω 1 = ? \log_2 |1+ \omega + \omega^2 + \omega^3 - \omega^{-1} | = \, ?


The answer is 1.

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1 solution

Soumya Shrivastva
Jan 13, 2016

ω 5 = 1 , 1 + ω + ω 2 + ω 3 + ω 4 = 0 , ω = 1 log 2 ω + ω 2 + ω 3 + ω 4 1 ω = log 2 2 ω = log 2 2 ω = log 2 2 1 = 1 { \omega }^{ 5 }=1,\quad 1+\omega +{ \omega }^{ 2 }+{ \omega }^{ 3 }+{ \omega }^{ 4 }=0,\quad \left| \omega \right| =1\\ \log _{ 2 }{ \left| \frac { \omega +{ \omega }^{ 2 }+{ \omega }^{ 3 }+{ \omega }^{ 4 }-1 }{ \omega } \right| } =\log _{ 2 }{ \left| \frac { -2 }{ \omega } \right| } =\log _{ 2 }{ \frac { \left| -2 \right| }{ \left| \omega \right| } } =\log _{ 2 }{ \frac { 2 }{ 1 } =1 }

How do you justify changing the sign of the last term, -1/omega ?

Tom Capizzi - 4 years, 8 months ago

This is only true if omega is a complex fifth root of unity. Omega = 1 is also a fifth root of unity. The equation simplifies to log2 3 <> 1.

Tom Capizzi - 4 years, 8 months ago

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