# 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.

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You can start with the factorization $t^5+1=(t+1)(1-t+t^2-t^3+t^4)$ –  Guess who it is. Apr 29 '12 at 10:45
Put $t=-s$, then $t^5+1=-(s^5-1)$. The roots of the latter are the $5^{\text{th}}$ roots of unity (four of which are complex), and so the roots of your equation are $-1$ times the $5^{\text{th}}$ roots of unity. Equations involving roots of unity are best solved using polar coordinates: see Gadi's answer below. –  Clive Newstead Apr 29 '12 at 11:00
@CliveNewstead, so the roots are -1, -w, -w^2, -w^3, -w^4, with w=e^(2ipi/5)? –  user29553 Apr 29 '12 at 11:11
Another way of arriving at Clive's result is to observe that $$t^5+1=\frac{t^{10}-1}{t^5-1},$$ so the roots of your equation are those tenth roots of unity that aren't fifth roots of unity. –  Jyrki Lahtonen Apr 29 '12 at 11:15
@user29553: Yes, that's right. You could also write $-1=e^{i\pi}$ so that instead of writing $-1,-\omega, \cdots,-\omega^4$ you have $e^{i\pi/5}, e^{3i\pi/5}, \cdots, e^{9i\pi/5}$ (which may or may not look neater depending on what you're doing). –  Clive Newstead Apr 29 '12 at 12:26

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).

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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:

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

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Thanks for the follow-through. :) Will upvote when I can. –  Guess who it is. Apr 30 '12 at 17:25