Real or Imaginary? Solving x^6 + 1

[Miike012]

My friends teacher posted a sample problem saying find the real roots of x^6 + 1. Is this a trick question? All roots for this function are imaginary right?

 

[tjackson3]

Yep!

Consider the fact that x^6 + 1 = 0 \Rightarrow x^6 = -1 \Rightarrow x = (-1)^{1/6}. If you write -1 = e^{i(\pi + 2\pi n)}, then this is simply e^{i(\pi/6 + n\pi/3)} = \cos(\pi/6 + n\pi/3) + i\sin(\pi/6 + n\pi/3) for n = 0, 1, ..., 5. If there were any real roots, then the imaginary part of this would be zero; i.e., we would have \pi/6 + n\pi/3 = m\pi for some integer m. But since there are only finitely many values of n, you can just do this manually. We have:

\pi/6, \pi/2, 5\pi/6, 7\pi/6, 3\pi/2, 11\pi/6

none of which would cancel the imaginary part.

 

[tjackson3]

If you've taken calculus (and I assume you have if you're on this board), another way to see it would be to note that since this is an even degree polynomial, the limit as x\rightarrow \pm\infty will be the same. In this case, it is positive infinity. Now find the minimum of this function by taking the derivative (6x^5) and setting it equal to zero. Then you can see that the only critical point is at x = 0, which corresponds to f(0) = 1 in the original function. You can perform a derivative test if you want to verify that it's a minimum, but graphically you can see that it is. Since the minimum is y = 1, there are no roots.

 

[HallsofIvy]

Well, not "imaginary", but "complex". the only "imaginary" roots of x^6+ 1= 0 is are i and -i. The other for are non-real complex numbers.