Roots of unity strike again

Algebra Level 4

ω − a 1 ω 2 − a 1 × ω − a 2 ω 2 − a 2 × ω − a 3 ω 2 − a 3 × ω − a 4 ω 2 − a 4 \frac { \omega-a_1 }{\omega^2-a_1 }\times\frac { \omega-a_2 }{ \omega^2-a_2 }\times \frac { \omega-a_3 }{ \omega^2-a_3 }\times\frac { \omega-a_4 }{ \omega^2-a_4 }

If 1 , a 1 , a 2 , a 3 , a 4 1,a_1,a_2,a_3,a_4 are the 5th roots of unity and ω \omega is a primitive ( ω ≠ 1 ) (\omega\neq1) 3rd root of unity, find the value of the expression above.

ω 2 \omega^2 ω \omega 1

Sanchayan Dutta by Sanchayan Dutta , 23, India

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3 solutions

Chew-Seong Cheong
Sep 21, 2015

( ω − a 1 ) ( ω − a 2 ) ( ω − a 3 ) ( ω − a 4 ) ( ω 2 − a 1 ) ( ω 2 − a 2 ) ( ω 2 − a 3 ) ( ω 2 − a 4 ) = ( ω 2 − 1 ) ( ω − 1 ) ( ω − a 1 ) ( ω − a 2 ) ( ω − a 3 ) ( ω − a 4 ) ( ω − 1 ) ( ω 2 − 1 ) ( ω 2 − a 1 ) ( ω 2 − a 2 ) ( ω 2 − a 3 ) ( ω 2 − a 4 ) = ( ω 2 − 1 ) ( ω 5 − 1 ) ( ω − 1 ) ( ω 10 − 1 ) 1 , a 1 , a 2 , a 3 , a 4 , fifth roots of unity. = ( ω 2 − 1 ) ( ω 2 − 1 ) ( ω − 1 ) ( ω − 1 ) ω 3 = 1 = ( ω + 1 ) ( ω + 1 ) ω 2 − 1 = ( ω + 1 ) ( ω − 1 ) = ω 2 + 2 ω + 1 ω 2 + ω + 1 = 0 = ω \begin{aligned} \frac{(\omega - a_1)(\omega - a_2)(\omega - a_3)(\omega - a_4)}{(\omega^2 - a_1)(\omega^2 - a_2)(\omega^2 - a_3)(\omega^2 - a_4)} & = \frac{(\omega^2 - 1) \color{#3D99F6}{(\omega - 1)(\omega - a_1)(\omega - a_2)(\omega - a_3)(\omega - a_4)}}{(\omega - 1) \color{#3D99F6}{(\omega^2 - 1)(\omega^2 - a_1)(\omega^2 - a_2)(\omega^2 - a_3)(\omega^2 - a_4)}} \\ & = \frac{(\omega^2 - 1) \color{#3D99F6}{(\omega^5 - 1)}}{(\omega - 1) \color{#3D99F6}{(\omega^{10} - 1)}} \quad \quad \small \color{#3D99F6}{1, a_1, a_2, a_3, a_4, \text{ fifth roots of unity.}} \\ & = \frac{(\omega^2 - 1)(\omega^2 - 1)}{(\omega - 1)(\omega - 1)} \quad \quad \small \color{#3D99F6} {\omega^3 = 1} \\ & = (\omega+1)(\omega +1 ) \quad \quad \quad \color{#3D99F6}{\omega^2-1=(\omega+1)(\omega-1)} \\ & = \omega^2 + 2\omega + 1 \quad \quad \quad \quad \color{#3D99F6}{\omega^2 + \omega+1 = 0} \\ & = \boxed{\omega} \end{aligned}

12 Helpful 0 Interesting 1 Brilliant 0 Confused
Alan Yan
Sep 22, 2015

( z − a 1 ) ( z − a 2 ) ( z − a 3 ) ( z − a 4 ) = z 4 + z 3 + z 2 + z + 1 (z - a_1)(z - a_2)(z - a_3)(z - a_4) = z^4 + z^3 + z^2 + z + 1

You know that ω 3 = 1 \omega^3 = 1 ω 2 + ω + 1 = 0 \omega^2 + \omega + 1 = 0

( ω − a 1 ) ( ω − a 2 ) ( ω − a 3 ) ( ω − a 4 ) ( ω 2 − a 1 ) ( ω 2 − a 2 ) ( ω 2 − a 3 ) ( ω 2 − a 4 ) = ω 4 + ω 3 + ω 2 + ω + 1 ω 8 + ω 6 + ω 4 + ω 2 + 1 \frac{(\omega - a_1)(\omega - a_2)(\omega - a_3)(\omega- a_4)}{(\omega^2 - a_1)(\omega^2 - a_2)(\omega^2 - a_3)(\omega^2- a_4)} = \frac{\omega^4 + \omega^3 + \omega^2 + \omega + 1}{\omega^8 + \omega^6 + \omega^4 + \omega^2 + 1}

= ω + ω ( ω 2 + ω + 1 ) + 1 ω 2 + 1 + ( ω + ω 2 + 1 ) = ω + 1 ω 2 + 1 ⋅ ω ω = ω ( ω + 1 ) 1 + ω = ω = \frac{\omega + \omega(\omega^2+\omega+1) + 1}{\omega^2 + 1 + (\omega + \omega^2 + 1)} = \frac{\omega+1}{\omega^2 + 1} \cdot \frac{\omega}{\omega} = \frac{\omega(\omega + 1)}{1 + \omega} = \boxed{\omega}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Lu Chee Ket
Oct 17, 2015

Excel method allows one to see exact orientations of triangle and pentagon:

w = -0.5+0.866025403784439i

w^2 = -0.5-0.866025403784438i

a1 = 0.309016994374947+0.951056516295154i

a2 = -0.809016994374947+0.587785252292473i

a3 = -0.809016994374948-0.587785252292473i

a4 = 0.309016994374947-0.951056516295154i

F1 = 0.204488531107294-0.354184525442962i

F2 = -0.13988638801609+0.242290331331161i

F3 = -1.78716459510874+3.09545988021662i

F4 = 1.22256245201755-2.11754028232038i

P = -0.5+0.866025403784439i

P = w TRUE

Chance of making mistake is lowest.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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