Complex roots of Unity Question (1 Viewer)

[cutemouse]

Hi all,

I have questions that are driving me CRAZY, I hope that someone could explain WHY it is this way (for 2nd question)...

Question are:

1. "If ὦ is a non-real fifth root of 1, show that 1+ὦ+ὦ²ὦ³ὦ⁴=0" [I have some clue on how to do this one, but it isn't driving me crazy]

2. "Factorise z⁵-1 as real factors... [this one I have NO clue, it's spinning my head] My working so far for this is below...

Now, for k=1 I have z₁=cis(4pi/5)

For k=-1 I have z₂=cis(-4pi/5)

I know that they're conjugate pairs, but when you add them together WHY do you get 2cis(4pi/5)??????? This is driving me CRAZY!!!

Isn't 4pi/5 in the 3rd quadrant, therefore cos should be negative, and therefore I should get -2cis(4pi/5)?? In other words, why does cis(-4pi/5) become cis(4pi/5)?

Thank you heaps if you can help me.

Jason

 

[adnan91]

u didnt complete the question..

 

[DownInFlames]

isn't 2pi/5 in the 1st quadrant? cause it's less than pi/2

 

[cutemouse]

Okay editted, sorry for the error, clearly I've been driven crazy.

Thanks,

Jason

 

[tommykins]

回复: Complex roots of Unity Question

Hi all,

I have a question that's driving me CRAZY, I hope that someone could explain WHY it is this way...

Question is: "If ὦ is a non-real fifth root of 1, show that 1+ὦ+ὦ²ὦ³ὦ⁴=0"

Now, for k=1 I have z₁=cis(4pi/5)

For k=-1 I have z₂=cis(-4pi/5)

I know that they're conjugate pairs, but when you add them together WHY do you get 2cis(4pi/5)??????? This is driving me CRAZY!!!

Isn't 4pi/5 in the 3rd quadrant, therefore cos should be negative, and therefore I should get -2cis(4pi/5)?? In other words, why does cis(-4pi/5) become cis(4pi/5)?

Thank you heaps if you can help me.

Jason

Lol, sum of roots = 0.

 

[cutemouse]

Sorry, editted again. Thanks alot if you can help me.

Thanks,

Jason

 

[tommykins]

回复: Complex roots of Unity Question

Hi all,

I have questions that are driving me CRAZY, I hope that someone could explain WHY it is this way (for 2nd question)...

Question are:

1. "If ὦ is a non-real fifth root of 1, show that 1+ὦ+ὦ²ὦ³ὦ⁴=0" [I have some clue on how to do this one, but it isn't driving me crazy]

2. "Factorise z⁵-1 as real factors... [this one I have NO clue, it's spinning my head] My working so far for this is below...

Now, for k=1 I have z₁=cis(4pi/5)

For k=-1 I have z₂=cis(-4pi/5)

I know that they're conjugate pairs, but when you add them together WHY do you get 2cis(4pi/5)??????? This is driving me CRAZY!!!

Isn't 4pi/5 in the 3rd quadrant, therefore cos should be negative, and therefore I should get -2cis(4pi/5)?? In other words, why does cis(-4pi/5) become cis(4pi/5)?

Thank you heaps if you can help me.

Jason

(z-1) is a given.
You obtain all the roots, and multiply them together in the form of (z+root) here. This should then cancel out the conjugate pairs leaving (z^2+2Re(z) + 1) or something like that, I forgot.

 

[Trebla]

Hi all,

I have a question that's driving me CRAZY, I hope that someone could explain WHY it is this way...

Question is: "If ὦ is a non-real fifth root of 1, show that 1+ὦ+ὦ²ὦ³ὦ⁴=0"

Now, for k=1 I have z₁=cis(4pi/5)

For k=-1 I have z₂=cis(-4pi/5)

I know that they're conjugate pairs, but when you add them together WHY do you get 2cis(4pi/5)??????? This is driving me CRAZY!!!

Isn't 4pi/5 in the 3rd quadrant, therefore cos should be negative, and therefore I should get -2cis(4pi/5)?? In other words, why does cis(-4pi/5) become cis(4pi/5)?

Thank you heaps if you can help me.

Jason

cis (2π/5) + cis (-2π/5) is NOT 2cis (2π/5), it is 2cos (2π/5)
cis (-2π/5) = cos (-2π/5) + isin (-2π/5)
Now using: sin (-x) = - sin x and cos (-x) = cos x
=> cis (-2π/5) = cos (2π/5) - isin (2π/5)
This is the conjugate of cis (2π/5) = cos (2π/5) + isin (2π/5)
[notice the sign of the imaginary part is different, so the two are in the form x + iy and x - iy]
When you add them up:
cis (2π/5) + cis (-2π/5)
= cos (2π/5) + isin (2π/5) + cos (2π/5) - isin (2π/5)
= 2cos (2π/5)

Second of all, with regards to the original question, it can be proved in a few lines without considering the actual roots themselves.
Since ω⁵ = 1 => ω⁵ - 1 = 0
But: (ω⁵ - 1) / (ω - 1) = (ω⁴ + ω³ + ω² + ω + 1) by geometric series formula, so
(ω⁵ - 1) = (ω - 1)(ω⁴ + ω³ + ω² + ω + 1)
=> (ω - 1)(ω⁴ + ω³ + ω² + ω + 1) = 0
But since ω is not real, then ω =/= 1, thus the other factor has to equal zero i.e.:
ω⁴ + ω³ + ω² + ω + 1 = 0

 

[cutemouse]

cis (2π/5) + cis (-2π/5) is NOT 2cis (2π/5), it is 2cos (2π/5)

Yeah sorry about that, I've been driven crazy... I mean 2cos

How about cos(-4π/5)? Since it's in 3rd quadrant, shouldn't cos be negative and thus give -2cos? I know it doesn't but this is the bit that I don't understand.

Thank you alot for your help, I really do appreciate it.

Thanks,

Jason

 

[cutemouse]

Hi again,

I guess what I really don't get is, how cos-4π/5 = cos4π/5, and sin-4π/5 = -sin4π/5

With -2π/5, I can understand it, because cos is positive in 4th quadrant, whilst sin is not and thus becomes negative, but for -4π/5, both sin and cos are in the 3rd quadrant, so shouldn't they both be negative? (This is the part driving me skits)

Thanks again,

Jason

 

[Trebla]

How about cos(-4π/5)? Since it's in 3rd quadrant, shouldn't cos be negative and thus give -2cos? I know it doesn't but this is the bit that I don't understand.

The theorem: sin (-x) = - sin x and cos (-x) = cos x holds regardless of how big or negative x actually is:
sin (-x) = - sin x because y = sin x is an odd function for ALL real x
cos (-x) = cos x because y = cos x is an even function for ALL real x
cis (-4π/5) + cis (4π/5) = 2cos (4π/5) < 0, since cos (4π/5) < 0
So yes, it is negative, but there is no negative sign because it's incoporated in the unevaluated cos (4π/5).

 

[cutemouse]

Hi,

Thank you very much for that.

Can you tell me what that theorem is called? I want to look it up in my textbook books, I don't think I found that it directed stated that.

Thanks,

Jason

 

[Trebla]

I don't think there's a name for it. It's merely common knowledge: (probably covered in Preliminary Trigonometry)

It arises from the fact that for an odd function f(-x) = - f(x), and for an even function f(-x) = f(x).

From the graph, you should realise that f(x) = sin x is an odd function (point symmetry about the origin), so
sin (-x) = - sin x

Similarly from the graph, you should realise that f(x) = cos x is an even function (symmetry about the y-axis), so
cos (-x) = cos x

 

[cutemouse]

Alright. Thank you, much appreciated

 

[chaldoking]

cis (2π/5) + cis (-2π/5) is NOT 2cis (2π/5), it is 2cos (2π/5)
cis (-2π/5) = cos (-2π/5) + isin (-2π/5)
Now using: sin (-x) = - sin x and cos (-x) = cos x
=> cis (-2π/5) = cos (2π/5) - isin (2π/5)
This is the conjugate of cis (2π/5) = cos (2π/5) + isin (2π/5)
[notice the sign of the imaginary part is different, so the two are in the form x + iy and x - iy]
When you add them up:
cis (2π/5) + cis (-2π/5)
= cos (2π/5) + isin (2π/5) + cos (2π/5) - isin (2π/5)
= 2cos (2π/5)

Second of all, with regards to the original question, it can be proved in a few lines without considering the actual roots themselves.
Since ω⁵ = 1 => ω⁵ - 1 = 0
But: (ω⁵ - 1) / (ω - 1) = (ω⁴ + ω³ + ω² + ω + 1) by geometric series formula, so
(ω⁵ - 1) = (ω - 1)(ω⁴ + ω³ + ω² + ω + 1)
=> (ω - 1)(ω⁴ + ω³ + ω² + ω + 1) = 0
But since ω is not real, then ω =/= 1, thus the other factor has to equal zero i.e.:
ω⁴ + ω³ + ω² + ω + 1 = 0

JIBRISH LOL - IS THAT FOR ONE SINGLE QUESTION :|

 

[Crates]

JIBRISH LOL - IS THAT FOR ONE SINGLE QUESTION :|

AHAHAHA, totally agree. That's absolute gibberish.

 

[tommykins]

JIBRISH LOL - IS THAT FOR ONE SINGLE QUESTION :|

Yes, it's not as complex (lolpun) as you think.

 

[Trebla]

JIBRISH LOL - IS THAT FOR ONE SINGLE QUESTION :|

It's gibberish to you, because you don't understand one bit of it

 

[chaldoking]

It's gibberish to you, because you don't understand one bit of it

I sure don't hehe. LOL at the 'pun' intended bit haha.