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    The question is {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}}

    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.
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    (Original post by Quido)
    The question is {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}}
    Should that be 1-w^6?
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    (Original post by Quido)
    The question is {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}}

    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}}?? It gives 2 different ones depending on which x you pick, unless you meant this part to be something different.
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    (Original post by atsruser)
    Should that be 1-w^6?
    Nope, the 6 is outside the bracket unless it is a misprint
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    (Original post by RDKGames)
    "The" value of {(1 - w)^{6}}?? It gives 2 different ones depending on which x 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.
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    (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-w is given by 1-e^{ix} for x \in \{ 0, \frac{2}{3}\pi , -\frac{2}{3}\pi \}

    Then you just consider each case separately.

    For 1-e^{\pm \frac{2}{3}\pi i} it would be useful to express this in cartesian form first before expressing it in the form re^{i\theta} then raising this to the 6th power.
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    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 Re^{i\ttheta} form without too much work.
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    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-e^{ix} = e^{ix/2} (e^{-ix/2} - e^{ix/2}) and the term in brackets can be rewritten as -2i sin(x/2).
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    (Original post by DFranklin)

    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:
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    (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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