How can I systematically find the roots of $ x^4 + 1?$ Is there some algorithm?

Question

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}

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? — , Dec 22 '12 at 13:44
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 $-1$. — Michael Hardy, Dec 22 '12 at 20:16

lab bhattacharjee · Answer 1 · 2012-12-22T13:17:49.837

3

$x^4=-1=e^{i\pi}$ (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 $0\le n<4$

edited Dec 22 '12 at 13:17

answered Dec 22 '12 at 13:06

lab bhattacharjee

2

If you can understand this answer, you wouldn't have needed it in the first place. – akkkk Dec 22 '12 at 13:08
@akkkk, if this comment is for me, I would request to explain the statement. Btw, did you down-vote? – lab bhattacharjee Dec 22 '12 at 13:11
How is someone who doesn't know how to find roots of $x^4+1$ helped by your answer? – akkkk Dec 22 '12 at 13:11
@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? – lab bhattacharjee Dec 22 '12 at 13:15
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. – akkkk Dec 22 '12 at 13:16
Better. Although I'd simply expect this question to be closed as duplicate. – akkkk Dec 22 '12 at 13:19
I do second you. – lab bhattacharjee Dec 22 '12 at 13:20

score 3 · Answer 2 · edited May 13 '14 at 08:52

3

$$x^4 + 1 = (x^2 + 1)^2 - 2x^2 = \\ (x^2+1+\sqrt2 x) (x^2+1-\sqrt2 x) =\\ (x^2 - x\sqrt 2+1)(x^2 + x\sqrt2+1)=0$$

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

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

edited May 13 '14 at 08:52

answered Dec 22 '12 at 13:28

Adi Dani

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." – amWhy Dec 22 '12 at 14:25
@amWhy. OK thanks – Adi Dani Dec 22 '12 at 14:28

2 Answers2