Fight with Omega

Algebra Level 5

Consider the equation x 5 − 1 = 0 x^5-1=0 . Suppose 1 , α , β , γ , δ 1,\alpha,\beta,\gamma,\delta be its distinct roots. Find the value of ( ω − α ) ( ω − β ) ( ω − γ ) ( ω − δ ) ( ω 2 − α ) ( ω 2 − β ) ( ω 2 − γ ) ( ω 2 − δ ) . \frac{(\omega-\alpha)(\omega-\beta)(\omega-\gamma)(\omega-\delta)}{(\omega^2-\alpha)(\omega^2-\beta)(\omega^2-\gamma)(\omega^2-\delta)} .

where ω \omega is the non-real cube root of 1.

ω \omega 1 1 ω 2 \omega^2

Atul Kumar Ashish by Atul Kumar Ashish , 21, India

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1 solution

Chew-Seong Cheong
Jan 12, 2017

χ = ( ω − α ) ( ω − β ) ( ω − γ ) ( ω − δ ) ( ω 2 − α ) ( ω 2 − β ) ( ω 2 − γ ) ( ω 2 − δ ) = ( ω − 1 ) ( ω − α ) ( ω − β ) ( ω − γ ) ( ω − δ ) ( ω 2 − 1 ) ( ω 2 − 1 ) ( ω 2 − α ) ( ω 2 − β ) ( ω 2 − γ ) ( ω 2 − δ ) ( ω − 1 ) As x 5 − 1 = ( x − 1 ) ( x − α ) ( x − β ) ( x − γ ) ( x − δ ) = ( ω 5 − 1 ) ( ω 2 − 1 ) ( ω 10 − 1 ) ( ω − 1 ) Note that ω 3 = 1 = ( ω 2 − 1 ) ( ω 2 − 1 ) ( ω − 1 ) ( ω − 1 ) = ( ω − 1 ) 2 ( ω + 1 ) 2 ( ω − 1 ) 2 = ( ω + 1 ) 2 = ω 2 + 2 ω + 1 = ω 2 + ω + 1 + ω = 0 + ω = ω \begin{aligned} \chi & = \frac {(\omega - \alpha)(\omega - \beta)(\omega - \gamma)(\omega - \delta)}{(\omega^2 - \alpha)(\omega^2 - \beta)(\omega^2 - \gamma)(\omega^2 - \delta)} \\ & = \frac {{\color{#3D99F6}(\omega -1)}(\omega - \alpha)(\omega - \beta)(\omega - \gamma)(\omega - \delta){\color{#D61F06}(\omega^2 -1)}}{{\color{#D61F06}(\omega^2 -1)}(\omega^2 - \alpha)(\omega^2 - \beta)(\omega^2 - \gamma)(\omega^2 - \delta){\color{#3D99F6}(\omega -1)}} & \small \color{#3D99F6}\text{As }x^5 - 1 = (x -1)(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) \\ & = \frac {{\color{#3D99F6}(\omega^5 -1)}{\color{#D61F06}(\omega^2 -1)}}{{\color{#D61F06}(\omega^{10} -1)}{\color{#3D99F6}(\omega -1)}} & \small \color{#3D99F6} \text{Note that }\omega^3 = 1 \\ & = \frac {(\omega^2 -1)(\omega^2 -1)}{(\omega -1)(\omega -1)} \\ & = \frac {(\omega -1)^2(\omega +1)^2}{(\omega -1)^2} \\ & = (\omega +1)^2 \\ & = \omega^2 + 2\omega + 1 \\ & = {\color{#3D99F6}\omega^2 + \omega + 1} + \omega \\ & = {\color{#3D99F6}0} + \omega \\ & = \boxed{\omega} \end{aligned}

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How w²+w+1=0

genis dude - 4 years, 4 months ago

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ω 3 = 1 ω 3 − 1 = 0 ( ω − 1 ) ( ω 2 + ω + 1 ) = 0 \begin{aligned} \omega^3 & = 1 \\ \omega^3 - 1 & = 0 \\ (\omega -1)(\omega^2 + \omega + 1) & = 0 \end{aligned}

⟹ { ω = 1 real root ω 2 + ω + 1 = 0 complex roots \implies \begin{cases} \omega = 1 & \small \color{#3D99F6} \text{real root} \\ \omega^2 + \omega + 1 = 0 & \small \color{#3D99F6} \text{complex roots} \end{cases}

Chew-Seong Cheong - 4 years, 4 months ago

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