# Complex polynomial roots?

1. Apr 14, 2005

### Exulus

Hi guys, can anyone tell me how I would go about solving this equation? :

$x^5 = x$

Rearranging it gives:
$x^5 - x = 0$

But then I don't really know what to do next. I know just from looking at it and thinking about it that the roots should be x = 0, 1, -1, -i, i...but I need to be able to come to that conclusion rather than state it.

Thanks for any help :)

2. Apr 14, 2005

### Galileo

Well, you can clearly see x=0 is a solution, so for the other solutions, you can divide the equation by x.
Can you solve the remaining equation? (There's a general method for finding the n n-th roots of unity).

By the way, since you have all the solutions, you can factor the polynomial or so.

Last edited: Apr 14, 2005
3. Apr 14, 2005

### Exulus

ahh of course! That'll explain it. I did wonder as i was looking through my notes, we were taught how to find roots of unity but not this. Dividing by x makes that possible, cheers :)

4. Apr 21, 2005

### abia ubong

hey
x can also be infinity

5. Apr 21, 2005

### Anzas

you can simply factor it like this:

$$x^5 - x = 0$$
$$x(x^4-1)=0$$
$$x(x^2+1)(x^2-1)=0$$
$$x(x^2+1)(x+1)(x-1)=0$$

now its obvious the real solutions are:
$$0,-1,1$$

and the complex solutions are:
$$x^2+1=0$$
$$x^2=-1$$
$$x=i$$
$$x=-i$$

Last edited: Apr 21, 2005
6. Apr 21, 2005

### eNathan

The only way this is posible is if $$x:=0; 0^5 = 0;$$

7. Apr 21, 2005

### dextercioby

Of course,the complex exponential is multivalued,that means that the # of solutions to

$$x^{4}=1$$

is infinite.

Daniel.

8. Apr 21, 2005

### abia ubong

nice work anzas,
and u exulus,my idea was this divide both side by x,u get x^4=1call this equation one,and rem. x[x^4-1]=0 this was resolved from the above question sub. x^4as 1 x(0)=0now divide both sides by 0 x=0/0 which is infinity.if u disagree let me know .see ya

9. Apr 23, 2005

### Data

bad idea. What does it mean to divide by zero exactly? And why do you think 0/0 is infinity?

I always thought the complex exponential was pretty single-valued. What's a value of $e^{ i\alpha}$, other than $\cos \alpha + i\sin \alpha$?

10. Apr 24, 2005

### mathwonk

the complex exponential is single valued. maybe he meant the complex log.