Factorising over C

[da_silvy]

x^4+x^3+x^2+x+1

D:?

[sql_lall]

neat trick #7218: Multiply by (x-1) ;)

also, by now that should look familiar to you as a roots-of-unity, Question. My hint: there are FIVE terms.

[da_silvy]

hmm

aren't there only complex factors (given that none of the coefficients are complex)

and aren't there only 4 factors - from the conjugate theorem?

[fundu]

As far as I think, there are no real roots of your function. But there are 4 complex roots.

I was just wondering what do you want to know?

1. Do you want an algorithm to tell U how many real and complex roots are there?
2. Do you want to find real roots?
3. Do you want to find both the real and complex roots?

What portion you want computer to do?

[da_silvy]

well seeing this is a maths forum i don't want the computer to do anything.

i don't see how this is hard!

Clearly I already know there are 4 factors and they are complex

I'm asking how to do it! (and what the solution is :p)

Thanks

[sql_lall]

For those who didn't get my tips, here's what they meant:

1) "Multiply by (x-1)"
(x-1)(x^4 + x^3 + x^2 + x^1) = x^5 - 1 :D:D -look familiar
=> Any root of of (x^4 + x^3 + x^2 + x^1) is also a root of x^5 - 1.
So, the roots u r looking for are the 4 complex fifth roots of unity.

2) "there are FIVE terms"
Looking at the equation for the nth roots of unity:
x^n - 1 = 0. factoring out the trivial root of x=1, gives:
x^n - 1 = (x-1)(x^(n-1) + x^(n-2) + .... + x + 1)

so, the roots of (x^(n-1) + x^(n-2) + .... + x + 1) are the nth roots of unity, not including the number 1.

Note that above there are n terms in that series, and you use the nth roots. because there are FIVE terms, the factors are the 5th roots of unity, not including x=1.

[da_silvy]

(x - cis (2pi / 5)) (x - cis (4pi / 5)) (x - cis (6pi / 5)) (x - cis (8pi / 5))

?

[sql_lall]

Thats right :D

Except i'd recommend doing it the standard way:
(x +/- cis(2Pi/5))(x +/- cis(4Pi/5))

That way u r shouwing u know about conjugates

[da_silvy]

cool, thanks :O)

[jskog]

Sql_lall, I enjoyed your solution to this problem. But what is CIS???? I've never seen that nomenclature before...

[da_silvy]

cis theta = cos theta + i sin theta

i.e. real part = Cos theta
imaginary part = Sin theta

[sql_lall]

Yeah, cis is short for cos+isin

Also, a better way to remember, is:

eix = cis(x) :D

this is using radians (...so ei.Pi + 1 = 0... ;))

[kedaman]

cis.. lol.. e^itheta for me, otherwise it gets confusing :D

[jskog]

Yeah. I totally agree with kedaman. Now I get what you were talking about. ;) In Sweden we always write
e^(i*theta)
:)