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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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    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

2 Answers 2

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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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    I do second $y$ou.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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    @amWhy. OK thanks2012-12-22