Roots of the unity

Algebra Level 3

Let a a be a complex number such that a 7 = 1 a^7 = 1 and a ≠ 1 a \neq 1 , and ω \omega be a primitive cubic root of the unity. What is the value of ∏ n = 1 7 ( ω − a n ) ∏ n = 1 7 ( ω 2 − a n ) + ω 2 − 1 ω 4 − 1 ? \large \frac{\displaystyle \prod_{n = 1}^7 (\omega - a^n)}{\displaystyle \prod_{n = 1}^7 (\omega^{2} - a^n)} + \frac{\omega^{2} - 1}{\omega^{4} - 1} ?

  • ω 2 \omega^{2}
ω \omega ω 2 \omega^{2} 1
  • ω \omega
-1 0

Guillermo Templado by Guillermo Templado , 46, Spain

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1 solution

Chew-Seong Cheong
Jan 29, 2018

X = ∏ n = 1 7 ( ω − a n ) ∏ n = 1 7 ( ω 2 − a n ) + ω 2 − 1 ω 4 − 1 Note that x 7 − 1 = ∏ n = 1 7 ( x − a n ) = ω 7 − 1 ω 14 − 1 + ω 2 − 1 ω − 1 Note that ω 3 = 1 = ω − 1 ω 2 − 1 + ( ω − 1 ) ( ω + 1 ) ω − 1 = ω − 1 ( ω − 1 ) ( ω + 1 ) + ω + 1 = 1 ω + 1 + ω + 1 Note that 1 + ω + ω 2 = 0 = 1 − ω 2 + ω + 1 = − ω 3 ω 2 + ω + 1 = − ω + ω + 1 = 1 \begin{aligned} X & = \frac {\color{#3D99F6}\prod_{n=1}^7 \left(\omega-a^n\right)}{\color{#3D99F6}\prod_{n=1}^7 \left(\omega^2-a^n\right)} + \frac {\omega^2-1}{{\color{#D61F06}\omega^4} -1} & \small \color{#3D99F6} \text{Note that } x^7 - 1 = \prod_{n=1}^7 \left(x-a^n\right) \\ & = \frac {\color{#3D99F6}\omega^7-1}{\color{#3D99F6}\omega^{14}-1} + \frac {\omega^2-1}{{\color{#D61F06}\omega} -1} & \small \color{#D61F06} \text{Note that } \omega^3 = 1 \\ & = \frac {\omega-1}{\omega^2-1} + \frac {(\omega-1)(\omega+1)}{{\omega} -1} \\ & = \frac {\omega-1}{(\omega-1)(\omega+1)} + \omega+1 \\ & = \frac 1{\color{#3D99F6}\omega+1} + \omega+1 & \small \color{#D61F06} \text{Note that } 1+\omega + \omega^2 = 0 \\ & = \frac 1{\color{#3D99F6}-\omega^2} + \omega+1 \\ & = - \frac {\omega^3}{\omega^2} + \omega+1 \\ & = - \omega + \omega+1 \\ & = \boxed{1} \end{aligned}

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