Cube Roots of Unity (1 Viewer)

[yummy-cookies]

I was wondering, how would you set up the proof for:

"If w is a cube root of unity (i. a root of z^3=1), prove that w^2 is also a complex cube root of unity."

I understand you need to use (z-1)(z^2+z+1)=0 to solve, but how would you actually set out the proof?

thanks

 

[lolokay]

you could just say

w³ = 1
(w³)² = 1
(w²)³ = 1
so w² is also a cube root of unity

 

[cutemouse]

Note that this stuff isn't examined much so I personally wouldn't worry about it.

 

[Aerath]

Awww, what? It's the bit of roots of unity that I do get. =P

 

[x.Exhaust.x]

Note that this stuff isn't examined much so I personally wouldn't worry about it.

It isn't examined much? Source?

And you're a HSC 2009 student man. I question the validity of your post .

Edited: 1500th post!

 

[cutemouse]

Alright then, why don't you go through 30 trial papers (from various years) and other HSC papers, like I did, and count how many times they've asked you to prove expressions using these results.

 

[jet]

Roots of unity are actually not that uncommon.
p.s i did my maths in 2008, so i do know.

 

[tacogym27101990]

yeah they do tend to pop up quite a bit
and its what most people struggle with the most in complex numbers.
that or vectors

 

[Timothy.Siu]

damn vectors

 

[cutemouse]

Roots of unity are actually not that uncommon.
p.s i did my maths in 2008, so i do know.

That's not what I was saying.

I was saying that in complex numbers they don't ask you to prove things using the results w³=1 and 1+w+w²=0 [WHICH IS THE TYPE OF QUESTION THAT THE OP IS ASKING]. I think it's still in the syllabus but I asked THREE maths teachers at my school (each with at least 25+ years experience) and they said to me not to worry about the proving expressions using those results, as it most likely will not be examined.

 

[gurmies]

Yeah, anybody else find vectors challenging? Not asking you lolokay, you're crazy!

 

[Aerath]

I find vectors hard too. =P

 

[gurmies]

Haha, yay, glad i'm not alone here!

 

[Trebla]

That's not what I was saying.

I was saying that in complex numbers they don't ask you to prove things using the results w³=1 and 1+w+w²=0 [WHICH IS THE TYPE OF QUESTION THAT THE OP IS ASKING]. I think it's still in the syllabus but I asked THREE maths teachers at my school (each with at least 25+ years experience) and they said to me not to worry about the proving expressions using those results, as it most likely will not be examined.

I disagree. If something hasn't been asked before, it has a much HIGHER chance of appearing in future exam papers. This is in reference to the harder questions of course at the latter half of the paper. You rarely ever find two or more papers that ask similar styles of questions in that section. Usually harder questions that have appeared in past HSC papers (particularly recent ones) don't generally appear again in future papers.

 

[kwabon]

damn vectors

ditto

 

[Tastegud]

you could just say

w³ = 1
(w³)² = 1
(w²)³ = 1
so w² is also a cube root of unity

That's true, but that's not really valid is it?
They would think: wow this guy is a smartass... WRONG

If you just show w³3 = "whatever" is a complex root, then find (w³)² = "whatever2"
and then "whatever"² SHOULD equal "whatever2"

Bad example tbh. I just relearnt this like a week ago, forgot all complex ==

 

[Trebla]

There is nothing wrong or invalid with what lolokay did.
Think about it, when you actually solve for complex roots of z³=1, you go:
Let z = w = rcis θ
w³ = 1
=> r³cis 3θ = cis (2kπ) for integer k
Equating modulus and argument:
r = 1, θ = 2kπ/3

If we sub in w² = r²cis 2θ for z³=1 instead of w = rcis θ
w⁶ = 1
=> (w²)³ = 1
=> r⁶ cis 6θ = cis (4kπ)
Equating modulus and argument:
r = 1, θ = 4kπ/6 = 2kπ/3
which is exactly identical to the previous case

 

[Aerath]

That's true, but that's not really valid is it?
They would think: wow this guy is a smartass... WRONG

If you just show w³3 = "whatever" is a complex root, then find (w³)² = "whatever2"
and then "whatever"² SHOULD equal "whatever2"

Bad example tbh. I just relearnt this like a week ago, forgot all complex ==

Why isn't it valid? I know it's the way I'd have done that question.

 

[jet]

I disagree. If something hasn't been asked before, it has a much HIGHER chance of appearing in future exam papers. This is in reference to the harder questions of course at the latter half of the paper. You rarely ever find two or more papers that ask similar styles of questions in that section. Usually harder questions that have appeared in past HSC papers (particularly recent ones) don't generally appear again in future papers.

Well said.

And to others, a lot of the questions involving roots of unity and trig involve the use of those identities that were mentioned earlier. I wouldn't jeopardise your performance in an exam just because you don't think it will be in it. 4 unit is a little different to 2 or 3 unit, you can't predict it that much.

 

[cutemouse]

Well said.

And to others, a lot of the questions involving roots of unity and trig involve the use of those identities that were mentioned earlier. I wouldn't jeopardise your performance in an exam just because you don't think it will be in it. 4 unit is a little different to 2 or 3 unit, you can't predict it that much.

Yes but I asked THREE maths teachers at my school who have been teaching for 25+ years and they don't feel that it will be examined, then I think they may be correct.

When do we use those results? I haven't come across and we've complex complex numbers and polynomials already.