The roots of $t^5+1$ Just a quick question, how do we go about finding the roots of $t^5 +1$?
I can see that since $t^5=-1$ that an obvious root is $\sqrt[5]{-1}$.
I am assuming that since there is a $-1$ involved, some of the factors will be complex?
Any help would be welcome.

Recall that you can represent every complex number using the so-called "exponential form": $a+bi=re^{i\theta}$ where $r=\sqrt{a^2+b^2}$ and $\tan\theta = \frac{b}{a}$ (when $a=0$, $\theta = \frac{\pi}{2}$ or $\theta = -\frac{\pi}{2}$, depending on the sign of $b$).
If this looks unfamiliar, note that it can be proved that $re^{i\theta}=r(\cos\theta+i\sin=\theta)$ which might be more familiar; this is the so-called "Euler's formula" - one of them).
This is an easy way to work with complex numbers when it comes to exponentiation, since $(re^{i\theta})^n=r^ne^{in\theta}$.
Now, $-1=e^{i\pi}$, so if $t=re^{i\theta}$ satisfies $t^5=-1$, we must have $r^5e^{i 5\theta}=1\cdot e^{i\pi}$. We get two equations:
$r^5=1$ which implies $r=1$ (since $r$ is a positive real number).
$e^{i 5\theta}=e^{i\pi}$ which implies $5\theta = \pi+2\pi k$ for every integer $k$, since this is the period of the trigonometric functions.
From this we get:
$\theta = \frac{\pi}{5}+\frac{2\pi}{5}k$
This is an infinite set of solutions, so we can restrict ourselves to those in the range $[0,2\pi]$ and end up with the solutions:
$\frac{\pi}{5},\frac{3\pi}{5},\pi,\frac{7\pi}{5},\frac{9\pi}{5}$.
Try to think how this can be generalized to root-finding of arbitrary numbers (solving general polynomial equations is much more difficult and can't be done with a nice equation for polynomials of degree 5 or more).
