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    Hi I need to solve

     z^4-z^2+1 as products of four factors in the form (z-e^{i\ {alpha}}) . Alpha is greater than or equal to 0 and less than 2pi.

    Is there a way to do this, without solving the quadratic, as it's not a very nice method.
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    What's not nice about it?

    The quartic is palindromic so you can write it as the product of two palindromic quadratics.

    z^4-z^2+1=(z^2+\sqrt{3} z +1)(z^2 - \sqrt 3 z +1)

    This goes easily but I'm not sure it isn't quicker just to solve the disguised quadratic directly.
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    (Original post by Mr M)
    What's not nice about it?

    The quartic is palindromic so you can write it as the product of two palindromic quadratics.

    z^4-z^2+1=(z^2+\sqrt{3} z +1)(z^2 - \sqrt 3 z +1)
    How did you work out what the middle term will be?

    I just solved the normal quadratic.
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    Thinking about it, why not just factorise the original cubic as:

    \displaystyle z^4-z^2+1=(z^2 - e^{\frac{i \pi}{3}})(z^2 - e^{-\frac{i \pi}{3}}) and take it from there?
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    (Original post by Mr M)
    Thinking about it, why not just factorise the original cubic as:

    \displaystyle z^4-z^2+1=(z^2 - e^{\frac{i \pi}{3}})(z^2 - e^{-\frac{i \pi}{3}}) and take it from there?
    I wouldn't know right of the bat, that it would factorize as that, how did you work it out? Also once I've solved the quadratic etc, and am readjusting the arguments, why is it + pi and not + 2pi?

    Thanks.
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    (Original post by Music99)
    How did you work out what the middle term will be?
    Seriously?

    The coefficient of z^2 needs to be -1.

    We have 1z^2 + 1z^2 so we need to subtract 3z^2.
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    (Original post by Mr M)
    Seriously?

    The coefficient of z^2 needs to be -1.

    We have 1z^2 + 1z^2 so we need to subtract 3z^2.
    Ah, my bad. It's getting late, and I'm having to teach myself FP3 and FP2 :P.
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    (Original post by Music99)
    Ah, my bad. It's getting late, and I'm having to teach myself FP3 and FP2 :P.
    Ah ok. I was assuming more knowledge than that. I thought this was an undergraduate thread - I've just noticed Sixth Form in the title.
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    (Original post by Mr M)
    Ah ok. I was assuming more knowledge than that. I thought this was an undergraduate thread - I've just noticed Sixth Form in the title.
    Well I am technically an undergrad, but transferring unis to York (fingers crossed), but have to do A-level further maths :P.
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    (Original post by Music99)
    I wouldn't know right of the bat, that it would factorize as that, how did you work it out? Also once I've solved the quadratic etc, and am readjusting the arguments, why is it + pi and not + 2pi?

    Thanks.
    I think the way you were supposed to approach this is to complete the square on the disguised quadratic.

    z^4 -z^2 +1=0

    (z^2 - \frac{1}{2})^2 - \frac{1}{4} + 1 = 0

    z^2 = \frac{1}{2} \pm \frac{i \sqrt{3}}{2}

    Now convert this pair of solutions to polar form and find the square roots to obtain four solutions and then write it as a set of four linear factors.
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    (Original post by Music99)
    Well I am technically an undergrad, but transferring unis to York (fingers crossed), but have to do A-level further maths :P.
    I see. Do you read mathematics at the moment? Where are you now?
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    (Original post by Mr M)
    I see. Do you read mathematics at the moment? Where are you now?
    I do, It's good just obviously some of the further maths stuff I haven't covered before. I'm at Brighton at the moment, and hopefully transferring to York.
 
 
 
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