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.
 A: 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).
A: 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:
$$ t \;\; = \;\; \frac{1}{4}\left( 1 - \sqrt{5}\right) \; \pm \; \left(\frac{1}{4} \sqrt{10 + 2\sqrt{5}}\right)i$$
