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Math Help - Cubic equation (complex numbers)

  1. #1
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    Cubic equation (complex numbers)

    Hi everyone, this is my first post here. If this belongs in a different section, my apologies.

    Question: P(z) := z^3 + pz + q, p and q are elements of Real numbers, u and v and Ϛ are elements of the complex numbers.

    u^3 = (-q/2) + sqrt((q/2)^2 + (p/3)^3) and v^3 = (-q/2) - sqrt((q/2)^2 + (p/3)^3)

    and Ϛ^3 = 1

    Show that z1 = u + v, z2 =
    Ϛu + (Ϛ^2)v and z3 = (Ϛ^2)u + Ϛv fufill the condition that P(z1) = P(z2) = P(z3) = 0

    I have the following:

    u^3 + v^3 = -q

    and (u*v)^3 = (u^3)*(v^3)
    = ((-q/2) + sqrt((q/2)^2 + (p/3)^3))*((-q/2) - sqrt((q/2)^2 + (p/3)^3))
    = ((-q^2)/4)) - ((p^3)/27)

    and

    P(z1) = (u+v)^3 + p(u+v) + q
    = u^3 + v^3 + 3((u^2)*v + u*(v^2)) + p(u+v) + q
    = 3((u^2)*v + u*(v^2)) + p(u+v)

    I don't know how to continue.

    Thanks for any help.
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  2. #2
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    Quote Originally Posted by Olym View Post
    Hi everyone, this is my first post here. If this belongs in a different section, my apologies.

    Question: P(z) := z^3 + pz + q, p and q are elements of Real numbers, u and v and Ϛ are elements of the complex numbers.

    u^3 = (-q/2) + sqrt((q/2)^2 + (p/3)^3) and v^3 = (-q/2) - sqrt((q/2)^2 + (p/3)^3)

    and Ϛ^3 = 1

    Show that z1 = u + v, z2 = Ϛu + (Ϛ^2)v and z3 = (Ϛ^2)u + Ϛv fufill the condition that P(z1) = P(z2) = P(z3) = 0

    I have the following:

    u^3 + v^3 = -q

    and (u*v)^3 = (u^3)*(v^3)
    = ((-q/2) + sqrt((q/2)^2 + (p/3)^3))*((-q/2) - sqrt((q/2)^2 + (p/3)^3))
    = ((-q^2)/4)) - ((p^3)/27)

    and

    P(z1) = (u+v)^3 + p(u+v) + q
    = u^3 + v^3 + 3((u^2)*v + u*(v^2)) + p(u+v) + q
    = 3((u^2)*v + u*(v^2)) + p(u+v)

    I don't know how to continue.

    Thanks for any help.
    Let's do it by parts: first, it's clear u^3 + v^3 = -q, so:

    P(z1) (u + v)^3 + p(u + v) + q = u^3 + v^3 + 3uv(u + v) + p(u + v) + q

    = (p + 3uv)(u + v).

    Now just check the first factor above p + 3uv is zero!

    With z2 and z3 is simmilar.

    Tonio
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  3. #3
    Super Member Matt Westwood's Avatar
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    Yes, this looks like an attempt to present Cardan's Formula.

    If you want to see a full exposition of his method of solving cubics, you can find it here:

    Cardan's Formula - ProofWiki
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  4. #4
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    Quote Originally Posted by Matt Westwood View Post
    Yes, this looks like an attempt to present Cardan's Formula.


    Don't make it french, he was italian: it is Gerolamo Cardano

    Tonio

    If you want to see a full exposition of his method of solving cubics, you can find it here:

    Cardan's Formula - ProofWiki
    -----------------------------------------------------------------
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  5. #5
    Super Member Matt Westwood's Avatar
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    Yes all right so he was Italian, but when I did the initial research, this formula was reported on Wikipedia as "Cardan's formula". I suppose it must have been updated since. (It's immaterial anyway, he wasn't actually the one who discovered it.)
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  6. #6
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    Quote Originally Posted by tonio View Post
    Let's do it by parts: first, it's clear u^3 + v^3 = -q, so:

    P(z1) (u + v)^3 + p(u + v) + q = u^3 + v^3 + 3uv(u + v) + p(u + v) + q

    = (p + 3uv)(u + v).

    Now just check the first factor above p + 3uv is zero!

    With z2 and z3 is simmilar.

    Tonio
    OK, unfortunately I don't understand what you mean by: "Now just check the first factor above p + 3uv is zero!".

    Another question (related to this one): If Ϛ^3 = 1 (and it's a complex number), then Ϛ^2 = 1 and Ϛ = 1, right?

    Thanks
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  7. #7
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Olym View Post
    Another question (related to this one): If Ϛ^3 = 1 (and it's a complex number), then Ϛ^2 = 1 and Ϛ = 1, right?

    Thanks
    Not necessarily. Consider \frac {-1 + i \sqrt 3}{2}. And of course its conjugate \frac {-1 - i \sqrt 3}{2}.
    Last edited by Matt Westwood; October 7th 2009 at 10:43 AM. Reason: Argh, grammar mistake!
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  8. #8
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    Quote Originally Posted by Olym View Post
    OK, unfortunately I don't understand what you mean by: "Now just check the first factor above p + 3uv is zero!".


    This means: prove that p + 3uv is zero...what else?

    Tonio



    Another question (related to this one): If Ϛ^3 = 1 (and it's a complex number), then Ϛ^2 = 1 and Ϛ = 1, right?

    Thanks
    .........................................
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  9. #9
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    Quote Originally Posted by Matt Westwood View Post
    Not necessarily. Consider \frac {-1 + i \sqrt 3}{2}. And of course it's conjugate \frac {-1 - i \sqrt 3}{2}.
    D'oh

    Quote Originally Posted by tonio View Post
    .........................................
    Sorry about that, I just misread what you wrote. It's been a long day...
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