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.

Since no one has yet pointed this out, it may be of interest to show how J.M.'s comment leads to a purely algebraic solution (i.e. no use made of trigonometric or exponential function ideas).
Using
$$t^5 + 1 \;\; = \;\; \left(t+1\right)\left(t^4 - t^3 + t^2 - t + 1\right),$$
it follows that the solutions to $t^5 + 1 = 0$ are $t=-1$ along with the solutions to
$$t^4 - t^3 + t^2 - t + 1 \;\; = \;\; 0$$
Dividing both sides of this last equation by $t^2$ gives
$$t^2 \; - \; t \; + \; 1 \; - \; \frac{1}{t} \; + \; \frac{1}{t^2} \;\; = \;\; 0$$
Rearranging terms gives
$$\left(t^2 \; + \; \frac{1}{t^2} \right) \; - \; \left(t \; + \; \frac{1}{t} \right) \; + 1 \; = \; 0$$
This is a reciprocal equation. (An equation with the property that if $t=r$ is a solution, then $t = \frac{1}{r}$ is also a solution. The phrase recurring equation was often used in the early and mid 1800s English literature.) Therefore, we make the substitution $u = t + \frac{1}{t}$ (note that $u^2 = t^2 + 2 + \frac{1}{t^2}$), which leads to
$$(u^2 - 2) \; - \; u \; + \; 1 \; = \; 0$$
$$u^2 - u - 1 \; = \; 0$$
Now use the quadratic formula to solve for $u$:
$$ u \;\; = \;\; \frac{-(-1) \; \pm \; \sqrt{(-1)^2 \; - \; 4(1)(-1)}}{2(1)} \;\; = \;\; \frac{1 \; \pm \; \sqrt{5}}{2}$$
This leads to the following two equations:
$$t \; + \; \frac{1}{t} \;\; = \;\; \frac{1 \; + \; \sqrt{5}}{2}$$
and
$$t \; + \; \frac{1}{t} \;\; = \;\; \frac{1 \; - \; \sqrt{5}}{2}$$
These two equations can be rewritten as:
$$2t^2 \; - \; \left(1 + \sqrt{5}\right)t \; + \; 2 \;\; = \;\; 0$$
and
$$2t^2 \; - \; \left(1 - \sqrt{5}\right)t \; + \; 2 \;\; = \;\; 0$$
Using the quadratic formula, the first of these equations has the following solutions:
$$ t \;\; = \;\; \frac{(1 + \sqrt{5}) \; \pm \; \sqrt{(6 + 2\sqrt{5}) \; - \; 16}}{4}$$
$$ t \;\; = \;\; \frac{(1 + \sqrt{5}) \; \pm \; i\sqrt{(10 - 2\sqrt{5}}}{4}$$
$$ t \;\; = \;\; \frac{1}{4}\left( 1 + \sqrt{5}\right) \; \pm \; \left(\frac{1}{4} \sqrt{10 - 2\sqrt{5}}\right)i$$
In the same way, the second of the these equations has the following solutions:
$$ t \;\; = \;\; \frac{1}{4}\left( 1 - \sqrt{5}\right) \; \pm \; \left(\frac{1}{4} \sqrt{10 + 2\sqrt{5}}\right)i$$
