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FP2 - Roots of Unity

The question is w=eix{w = e^{ix}} is a cube root of 1, state the three values of x in the range -pi < x < pi and find the possible values of (1−w)6{(1 - w)^{6}}

For the possible values of x, I have got 2pi/3, -2pi/3 and 0. I also know that 1 + w + w^2 = 0 although I don't know how to move from there.

Thanks in advance.
(edited 7 years ago)
Original post by Quido
The question is w=eix{w = e^{ix}} is a cube root of 1, state the three values of x in the range -pi < x < pi and find the value of (1−w)6{(1 - w)^{6}}

Should that be 1−w61-w^6?
Original post by Quido
The question is w=eix{w = e^{ix}} is a cube root of 1, state the three values of x in the range -pi < x < pi and find the value of (1−w)6{(1 - w)^{6}}

For the possible values of x, I have got 2pi/3, -2pi/3 and 0. I also know that 1 + w + w^2 = 0 although I don't know how to move from there.

Thanks in advance.


"The" value of (1−w)6{(1 - w)^{6}}?? It gives 2 different ones depending on which xx you pick, unless you meant this part to be something different.
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Original post by atsruser
Should that be 1−w61-w^6?


Nope, the 6 is outside the bracket unless it is a misprint :redface:
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Original post by RDKGames
"The" value of (1−w)6{(1 - w)^{6}}?? It gives 2 different ones depending on which xx you pick, unless you meant this part to be something different.


Sorry, I was meant to write the possible values of (1 - w)^6
Changed now.
Original post by Quido
Sorry, I was meant to write the possible values of (1 - w)^6
Changed now.


Ah that's better.

Okay so 1−w1-w is given by 1−eix1-e^{ix} for x∈{0,23π,−23π}x \in \{ 0, \frac{2}{3}\pi , -\frac{2}{3}\pi \}

Then you just consider each case separately.

For 1−e±23πi1-e^{\pm \frac{2}{3}\pi i} it would be useful to express this in cartesian form first before expressing it in the form reiθre^{i\theta} then raising this to the 6th power.
Well, one value for (1-w)^6 should be easy...

For the others, draw 1-w on an argand diagram. Some basic geometry should let you express 1-w in
Unparseable latex formula:

Re^{i\ttheta}

form without too much work.
PS: if you want to do it without an Argand diagram, I think you'll find life a bit easier if you observe that

1−eix=eix/2(e−ix/2−eix/2)1-e^{ix} = e^{ix/2} (e^{-ix/2} - e^{ix/2}) and the term in brackets can be rewritten as -2i sin(x/2).
(edited 7 years ago)
Original post by DFranklin


1−eix=eix/2(e−ix/2−eix/2)1-e^{ix} = e^{ix/2} (e^{-ix/2} - e^{ix/2}) and the term in brackets can be rewritten as -2i sin(ix/2).


Typo? :ninja:
Original post by RDKGames
Typo? :ninja:
Well, brain-fart...
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Alright, I've got it now. Thanks everyone!

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