Complex Cube root of Unity (1 Viewer)

[cutemouse]

Hi all,

I have a question, I'm mainly interested in the setting out of this question, so please do post full working.

"If ω is a complex cube root of unity, show that the other complex cube root is ω² and prove that 1+ω+ω²=0

Show that a) (1-3ω+ω²)(1+ω-8ω²)=36

b) (7+9ω⁴+7ω^-1)⁶=64

Thanks,

Jason

 

[Trebla]

Hi all,

I have a question, I'm mainly interested in the setting out of this question, so please do post full working.

"If ω is a complex cube root of unity, show that the other complex cube root is ω² and prove that 1+ω+ω²=0

Show that a) (1-3ω+ω²)(1+ω-8ω²)=36

b) (7+9ω⁴+7ω^-1)⁶=64

Thanks,

Jason

I assume you mean if ω is a non-real cube root of unity.
From z³ = 1,
Let z = rcis θ where r and θ are real
=> r³cis 3θ = cis 2πk for some integer k
Equating modulus and argument:
r³ = 1 => r = 1
3θ = 2πk => θ = 2πk/3
Therefore solutions are:
cis 0, cis 2π/3, cis 4π/3
Since ω is non-real, then we take either ω = cis 2π/3 or ω = cis 4π/3
If ω = cis 2π/3
ω² = cis 4π/3 by DeMoivre's theorem
which is the other root
If ω = cis 4π/3 instead
ω² = cis 8π/3 by DeMoivre's theorem
= cis (2π + 2π/3)
= cis 2π/3
which is the other root
Thus, if one cubic root of unity is ω then ω² is always the other root

As for proving 1 + ω + ω² = 0:
Since ω satisfies z³ = 1 then
ω³ = 1
ω³ - 1 = 0
(ω - 1)(ω² + ω + 1) = 0
If ω is non-real then: (ω² + ω + 1) = 0
Alternatively you could use sum of roots.

a) LHS = (1 - 3ω + ω²)(1 + ω - 8ω²)
= (1 + ω + ω² - 4ω)(1 + ω + ω² - 9ω²)
= (-4ω)(-9ω²) since (ω² + ω + 1) = 0
= 36ω³
= 36 since ω³ = 1
= RHS

b) LHS = (7 + 9ω⁴ + 7ω^-1)⁶
= (7 + 9ω³ω + 7/ω)⁶
= (7 + 9ω + 7/ω)⁶ as ω³ = 1
= (7ω + 9ω² + 7)⁶ / ω⁶
= (9ω² + 7ω + 7)⁶ / (ω³)²
= (7ω² + 7ω + 7 + 2ω²)⁶ as ω³ = 1
= (7(ω² + ω + 1) + 2ω²)⁶
= 2⁶ω¹² since (ω² + ω + 1) = 0
= 2⁶(ω³)⁴
= 64 as ω³ = 1
= RHS

 

[independantz]

Not quite sure what you mean "the other complex cube root"...

I believe he means solve z^3=1. Yo
u will get 3 roots one of these roots will be 1, the other will be cis(n/3) and the other will be cis(2n/3).
let w= cis(n/3)
consider: cis(2n/3)=[cis(n/3)}^2, by demoivre's thereom
therefore, cis(2n/3)=w^2

thus, w^2 is also a cube root of unity

note: n represents pie

 

[cutemouse]

Hi Trebla,

Thanks for your help. I have a few questions though, in the book (Coroneos) it says in the examples to not find the actual roots (unless specifically requested) but use the results of 1+ω+ω²=0 and ω³=1 (I think...) So...?

Also I don't get this bit below. Could you explain how cis (2π + 2π/3) = cis 2π/3 please?

ω² = cis 8π/3 by DeMoivre's theorem
= cis (2π + 2π/3)
= cis 2π/3
which is the other root

Thanks again for your help!

 

[Trebla]

Hi Trebla,

Thanks for your help. I have a few questions though, in the book (Coroneos) it says in the examples to not find the actual roots (unless specifically requested) but use the results of 1+ω+ω²=0 and ω³=1 (I think...) So...?

Also I don't get this bit below. Could you explain how cis (2π + 2π/3) = cis 2π/3 please?


Thanks again for your help!

cis (2π + 2π/3) = cis 2π/3 because
cos (2π + 2π/3) = cos (2π/3)
sin (2π + 2π/3) = sin (2π/3)
i.e. sin/cos (2π + x) = sin/cos x

If you do not find the actual roots, then this is what I came up with:
For the equation: z³ = 1 => z³ - 1 = 0
Since ω satisfies the equation then ω³ = 1
Now test if ω² works, sub into z³ - 1 = 0
LHS = z³ - 1
= (ω²)³ - 1 when subbing in z = ω²
= (ω³)² - 1
= 1 - 1 (since ω³ = 1)
= 0
= RHS
Thus, ω² is another root by the factor theorem.

 

[cutemouse]

Thanks alot for that, I think it is that method, without finding the actual roots.

But I have one more question, Hope you don't mind answering please:

cis (2π + 2π/3) = cis 2π/3 because
cos (2π + 2π/3) = cos (2π/3)
sin (2π + 2π/3) = sin (2π/3)
i.e. sin/cos (2π + x) = sin/cos x

Wouldn't cos be negative as it'd be in the 2nd quadrant if you add 2π?

Thanks,

Jason

 

[shaon0]

hey what are complex roots of unity? i've done some of complex numbers but haven't come across them yet. Are they hard?

 

[Trebla]

Thanks alot for that, I think it is that method, without finding the actual roots.

But I have one more question, Hope you don't mind answering please:


Wouldn't cos be negative as it'd be in the 2nd quadrant if you add 2π?

Thanks,

Jason

cos (2π/3) is indeed a negative number, but we've left it in its functional form without actually computing it's numerical value. The negative sign is incorporated within the cosine function (i.e. cos (2π/3) < 0).

I mean you wouldn't write cos (2π + 2π/3) = - cos (2π/3) because if you try it on your calculator, the RHS gives a positive number, whereas the LHS gives a negative number. (so the equality is false)

It's a bit like saying (1 - 2) is a negative number, but we haven't actually found its numerical value (in which case the negative sign does not appear unless we write it in simpler form as -1). The rule cos (2π + x) = cos x and sin (2π + x) = sin x applies regardless of whether x is positive, negative or zero.

 

[cutemouse]

Thanks alot, much appreciated!

The rule cos (2π + x) = cos x applies regardless of whether x is positive, negative or zero.

This rule is the same for sin?

Thanks,

Jason