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Math Help - DeMoivre's Theorem problem

  1. #1
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    DeMoivre's Theorem problem

    ...
    Last edited by harold; July 27th 2006 at 02:18 PM.
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    Quote Originally Posted by harold
    Using DeMoivre's Theorem (or a possible simpler method if there is one) find all the roots of x^4 + 3.

    Yikes!!
    I presume you can to factor this, based on your other post on irreducible polynomials?

    x^4+3=\sqrt[4]{3} (1-i)(1+i)(-1+i)(-1-i)
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    Quote Originally Posted by harold
    Hi TPH,

    Yes--I think I have to use this for my other post. What did you substitute in for demoivre's theorem?
    You have,
    x^4=-3
    In polar complex form,
    x^4=3(\cos \pi+i\sin \pi)
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    Quote Originally Posted by harold
    Hi TPH
    Could you run through how you got the factored version using the above? Also, in the other forum, I still don't see how to bring it all together to conclude irreducibility (as well as factor part b). Thanks in advance for any help.
    harold
    I am not going to do every single step, you are a big boy you can handle it yourself.

    As you understand (I hope),
    x^4=3(\cos \pi +i\sin \pi)
    In order to use de Moiver's theorem you write,
    x^4=3[\cos (\pi +2\pi k)+i\sin (\pi +2\pi k)]
    Now you use de Moiver's theorem (actually it is not cuz his theorem if for integral exponents not rational).
    Thus,
    x=3^{1/4}\left[\cos \left( \frac{\pi +2\pi k}{4} \right) +i\sin \left( \frac{\pi +2\pi k}{4} \right)\right]
    And each solution is for k=0,1,2,3
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    Quote Originally Posted by harold
    Isn't the above on the right of the = sign just a constant? How does it equal the left?

    I found the four solutions using k=0,1,2,3 from DeMoivre's Thm...how do I form these into a product of linear factors? I'm also having trouble doing this for part b) for my irreducible polys thread.
    Tell me your solutions, so I can see if you are doing it correctly.

    (Basically you use the factor theorem a is zero of f(x) if and only if (x-a) divides f(x))
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    Ignore what I said before I made a mistake.
    -------
    This is the factorization
    <br />
\left(<br />
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right) <br />
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right) <br />
\left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right) <br />
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)<br />
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  7. #7
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    Quote Originally Posted by harold
    And so now I can use the above as a guide to answer all my irreducible poly questions in the following way:

    Since <br />
\left(<br />
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right) <br />
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right) <br />
\left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right) <br />
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)<br />
    is written as a product of irreducibles in \mathbb C[x] (since each is degree 1) and
    <br />
\left(<br />
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right), <br />
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)<br />
are not in \mathbb R[x] but notice they are conjugates, so <br />
\left(<br />
x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right) <br />
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right) <br />
\left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right) <br />
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)<br />
=x^4+3 is in \mathbb R[x] and is irreducible in \mathbb R[x]
    That is the factorization in \mathbb{C}[x], good. Now find its factorization in \mathbb{R}[x]. The trick is to look for the conjugates. Remember that, (x-z_1)(x-z_2)=x^2-x(z_1+z_2)+z_1z_2, where z_1+z_2 is no longer complex and z_1z_2 is no longer complex.

    The first pair of conjugates are,
    <br />
\left( x- \sqrt[4]{3}[\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)\cdot \left(x-\sqrt[4]{3}[\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)<br />
    The polynomial they produce is,
    x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}
    The other pairs of conjugates are,
    <br />
\left( x- \sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} - i \frac {\sqrt {2}} {{2}}] \right)\cdot \left( x-\sqrt[4]{3}[-\frac {\sqrt{2}} {{2}} + i \frac {\sqrt {2}} {{2}}]\right)<br />
    The polynomial they produce is,
    x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}
    So the factorization of,
    x^4+3 in \mathbb{R}[x] is,
    [x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}][x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}]
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