Question

Find all the roots of the equation \(x^5-1=0\).

Answer 1

write 1^(1/5) as e^(i*2pi)/5
then take move around the unit circle to get the rest of the roots.

Answer 2

Is the answer 1?(only one root)

Answer 3

1 real root, 5 complex ones

Answer 4

5 or 4?

Answer 5

5, including the real one

Answer 6

Yes.

Answer 7

Isn't that the fundamental theorem of algebra?

Answer 8

idk what fundamental theorem of algebra is?

Answer 9

I though there was some kind of factorization rule for
x^n - a^n = (x-a)(....) ... nevertheless, the method by phi is far superior.

Answer 10

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

Answer 11

Oh, that was obvious. Degree is 5, so a total of 5 roots.

Answer 12

Can you find the other roots? (2pi/5+2pi), ...?

Answer 13

wolframalpha style? http://www.wolframalpha.com/input/?i=x%5E5-1%3D0

Answer 14

Yeah, I could have done that. :/ silly me. Thanks btw.

Answer 15

interesting.

Answer 16

.....

Answer 17

in case it is not clear, you move in increments of 2pi/5 : you get (as the exponent to e)
2pi/5 , 4pi/5 , 6pi/5, 8pi/5, 10pi/5
the last is just 2*pi and you are at 1 on the real axis

Answer 18

\[\large x^5-1 = 0\]
\[\large \implies (x - e^{\frac{i2\pi}{5}})(x - e^{\frac{i4\pi}{5}})(x - e^{\frac{i6\pi}{5}})(x - e^{\frac{i8\pi}{5}})(x - 1)=0\]
So, can I do this?
\[\int \frac{1}{x^5-1}dx = \frac{1}{(x - e^{\frac{i2\pi}{5}})(x - e^{\frac{i4\pi}{5}})(x - e^{\frac{i6\pi}{5}})(x - e^{\frac{i8\pi}{5}})(x - 1)}dx\]

Answer 19

if x^5-1 = 0 isn't this undefined?

Answer 20

The problem is to integrate the function \(\large\frac{1}{x^5-1}dx\).
\[x^5-1 = (x - e^{\frac{i2\pi}{5}})(x - e^{\frac{i4\pi}{5}})(x - e^{\frac{i6\pi}{5}})(x - e^{\frac{i8\pi}{5}})(x - 1)\]Maybe I can use this to integrate it.

Answer 21

Partial fractions. but I'm not sure.

Answer 22

I haven't tried anything like this before.

Answer 23

It looks like a lot of work. post your answer if you can get to one.

Answer 24

The answer on wolframalpha is pretty huge D: