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Possible Duplicate:
How to find the root of $x^4 +1$

What algorithms can be used for finding all roots of the given polynomial:

\begin{equation} x^4 + 1 = 0 \end{equation}

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    It depends somewhat on what you mean by "find" -- if the end result of 'finding' is the phrase "they are the primitive eighth roots of unity", is that good enough?2012-12-22
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    You get this factorization almost instantaneously if you know that multiplication by a non-zero complex number consists of rotating and dilating. Use that to find fourth roots of $-1$.2012-12-22

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$x^4=-1=e^{i\pi}$ (Using Euler's formula )

So, $x^4=e^{(2n+1)\pi i}$ where $n$ is any integer.

Using de Moivre's formula for fractional index, $x=e^{\frac{(2n+1)\pi i}4}$ where $0\le n<4$

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    If you can understand this answer, you wouldn't have needed it in the first place.2012-12-22
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    @akkkk, if this comment is for me, I would request to explain the statement. Btw, did you down-vote?2012-12-22
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    How is someone who doesn't know how to find roots of $x^4+1$ helped by your answer?2012-12-22
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    @akkkk, as one is helped by the accepted answer in http://math.stackexchange.com/questions/233999/how-to-find-the-root-of-x4-1. Btw, what was your expected answer?2012-12-22
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    Yes, those answers are much better and more welcoming. This just smacks an answer to the question, which is technically correct, but the techniques it relies on are not explained at all.2012-12-22
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    Better. Although I'd simply expect this question to be closed as duplicate.2012-12-22
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    I do second you.2012-12-22
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$$x^4 + 1 = (x^2 + 1)^2 - 2x^2 = \\ (x^2+1+\sqrt2 x) (x^2+1-\sqrt2 x) =\\ (x^2 - x\sqrt 2+1)(x^2 + x\sqrt2+1)=0$$

  1. $x^2 - x\sqrt 2+1=0$ $$x_{1}={\sqrt2+\sqrt{-2}\over2}={\sqrt2\over2}(1+i)$$ $$x_{2}={\sqrt2-\sqrt{-2}\over2}={\sqrt2\over2}(1-i)$$
  2. $x^2 + x\sqrt2+1=0$

$$x_{3}={-\sqrt2+\sqrt{-2}\over2}=-{\sqrt2\over2}(1-i)$$ $$x_{4}={-\sqrt2-\sqrt{-2}\over2}=-{\sqrt2\over2}(1+i)$$

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    Adi Dani - I simply adjusted the formatting so the two possibilities display as "1. " and "2. ". Your text indicated that, but it displayed as "1. " and "1."2012-12-22
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    @amWhy. OK thanks2012-12-22