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Math Help - Factoring z^4+1 over the complex numbers

  1. #1
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    Factoring z^4+1 over the complex numbers

    to do this problem you just set z= x+yi then just factor away right? i just need to know because homework is worth 40% of my grade in this class....ty very much!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Find the four roots z_0,z_1,z_2,z_3 of z^4=-1 i.e. z_j=\sqrt [4] {-1} then,

    z^4+1=(z-z_0)(z-z_1)(z-z_2)(z-z_3)
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  3. #3
    Junior Member lanierms's Avatar
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    It might be hard to find the four roots. Here is what it would look like if I do what FernandoRevilla said.

    Solution:

    z^4+1

    =z^4-i^2

    =(z^2+i)(z^2-i)

    =(z^2+i)(z^2-(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^2)

    =(z^2+i)(z+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)  (z-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)

    =(z^2-(-i))(z+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)(z-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)

    =(z^2-(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)^2)(z+\frac{\sqrt{2}}{2}+\frac  {\sqrt{2}}{2}i)(z-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)

    =(z+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)(z-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)(z+\frac{\s  qrt{2}}{2}+\frac{\sqrt{2}}{2}i)(z-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)

    Why is Latex so confusing.....
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  4. #4
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    Quote Originally Posted by lanierms View Post
    Solution:

    z^4+1

    =z^4-i^2

    =(z^2+i)(z^2-i)
    Why not at this step write z^2+i = 0 \Rightarrow z^2 = -i \Rightarrow z = \pm i\sqrt{i} or
    z^2-i \Rightarrow z^2 = i \Rightarrow z = \pm \sqrt{i} (or it's just the difference of two squares).
    (So the four roots of the given equation are \sqrt{i}, -\sqrt{i}, i\sqrt{i}, and -i\sqrt{i} ) Thus:
    z^4+1 = (z^2+i)(z^2-i) = (z-i\sqrt{i})(z+i\sqrt{i})(z-\sqrt{i})(z+\sqrt{i}).
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    An alternative:

    \sqrt[4]{-1}=\sqrt[4]{\cos \pi/4+i\sin \pi/4}=

    \left\{{z_k=\cos (\pi/4+k\pi/2)+i\sin (\pi/4+k\pi/2):k=0,1,2,3}\right\}=

    \left\{{\sqrt{2}/2(1+i),\sqrt{2}/2(-1+i),\sqrt{2}/2(-1-i),\sqrt{2}/2(1-i)}\right\}

    So,

    z^4+1=(z-z_0)(z-z_1)(z-z_2)(z-z_3)

    in the form lanierms provided i.e. separated real and imaginary parts of the roots.
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  6. #6
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    Note that ChrisYee08 said "homework is worth 40% of my grade in this class". It would have been better to give some hints rather than give the entire answer.
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    It would have been better to give some hints rather than give the entire answer.
    I agree, so I did im my first post. After the complete solution of lanierms it was irrelevant to add or not to add different alternatives.
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  8. #8
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    wow cool thanks. its ok if its just the solution. when i have the solution im always able to go back and figure out what i did wrong and how to do it correctly. i was jw how did you some up with roots?
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  9. #9
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    No, it's not okay! It is very easy to look at a solution given by someone else and tell yourself that you understand it now, but even if you do understand every step, you have not had the experience of finding the steps for yourself- and that is where learning occurs.
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