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Thread: Solving Quartic

  1. #1
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    Solving Quartic

    Find all roots of x^4 + i =0. Write answers in trigonometric form.

    Don't know how to solve quartics.
    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by c_323_h
    Find all roots of x^4 + i =0. Write answers in trigonometric form.

    Don't know how to solve quartics.
    Any help would be appreciated.
    You need to use a theorem know as De' Moivre's
    .
    You need to express
    $\displaystyle x^4+i=0$
    as,
    $\displaystyle x^4=-i$
    Now, express $\displaystyle i$ in trigonomteric form,
    $\displaystyle i=\cos\frac{-\pi}{2}+i\sin\frac{-\pi}{2}$
    Thus,
    $\displaystyle x^4=\cos\frac{-\pi}{2}+i\sin\frac{-\pi}{2}$
    Thus,
    $\displaystyle x=\left(\cos\frac{-\pi}{2}+i\sin\frac{-\pi}{2}\right)^{1/4}$
    By De Moivre's Theorem,
    $\displaystyle x=\cos\left(-\frac{\pi}{8}\right)+i\sin\left(-\frac{\pi}{8}\right)$
    But, this is only one solution there are 3 more (because of the degree). In this case, we divide, $\displaystyle 2\pi$ by the degree, (4) to get, $\displaystyle \pi/2$ and each other solution is added by $\displaystyle \pi/2$
    Thus,
    $\displaystyle x=\left(\cos\frac{-\pi}{8}+i\sin\frac{-\pi}{8}\right)$
    $\displaystyle x=\left(\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8} \right)$
    $\displaystyle x=\left(\cos\frac{7\pi}{8}+i\sin\frac{7\pi}{8} \right)$
    $\displaystyle x=\left(\cos\frac{11\pi}{8}+i\sin\frac{11\pi}{8} \right)$
    If you want you can add $\displaystyle 2\pi$ to your first solution to get,

    $\displaystyle x=\left(\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8} \right)$
    $\displaystyle x=\left(\cos\frac{7\pi}{8}+i\sin\frac{7\pi}{8} \right)$
    $\displaystyle x=\left(\cos\frac{11\pi}{8}+i\sin\frac{11\pi}{8} \right)$
    $\displaystyle x=\left(\cos\frac{15\pi}{8}+i\sin\frac{15\pi}{8} \right)$

    The beauty of the second result is that it is very easy to see, that each one is divided by 8 and the numerator is increasing by 4.
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    Thanks!

    when you solve for roots do you always solve for x?
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  4. #4
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    Quote Originally Posted by c_323_h
    Thanks!

    when you solve for roots do you always solve for x?
    Not sure I understand your question
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  5. #5
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    $\displaystyle x^2 + i = 0 $

    You subracted $\displaystyle i$ from both sides to get $\displaystyle x^4=-i$

    Why did you subtract $\displaystyle i$?
    Last edited by c_323_h; Apr 2nd 2006 at 03:46 PM.
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  6. #6
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    Quote Originally Posted by c_323_h
    $\displaystyle x^2 + i = 0 $

    You subracted $\displaystyle i$ from both sides to derive $\displaystyle x^4=-i$

    Why did you subtract $\displaystyle i$?
    You mean, in a regular poynomial you factor one side so you can set its factors equal to zero but over here you do not do this?

    Because a special equation of polynomial is,
    $\displaystyle x^n=y,y\not =0$.
    We can show that they are EXACTLY $\displaystyle n$ solutions and there are called Roots of Unity which are of quite importance in Abstract Algebra/Field Theory.

    Whenever, you see and equation like this you bring the constant term to the other side and then use De Moivre's Theorem to find its roots.
    -----
    It is interesting to note, that the soltuions that you get, are in trigonomteric form. Some of them can be simplified in algebraic terms (only using +,-,x,/ and roots) while some of them cannot (impossible) and thus you need to leave it in the form it is in.
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