# How can I find the fifth root of unity?

I need to find fifth root of unity in the form $x+iy$. I'm new to this topic and would appreciate a detailed "dummies guide to..." explanation!

I understand the formula, whereby for this question I would write: $1^{1/5} = r^{1/5}e^{2ki\pi/5}$. However, I don't know what to do next.

Any help is appreciated.

• Do you know the formula $e^{i\theta}=\cos \theta+i\sin \theta$? – LASV Dec 11 '13 at 21:24

Let's do it the hard way. We want to solve the equation $$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=0.$$ Then we are interested in solving $$x^4+x^3+x^2+x+1=0$$. Note the symmetry: if $$r$$ is a root, $$1/r$$ is also a root. Why? Thereafter, we have two ways out. First, we divide everything by $$x^2$$ to get $$x^2+x+1+\frac 1x+\frac{1}{x^2}=\left(x+\frac1x\right) +\left(x+\frac1x\right)^2-1=u^2+u-1=0.$$ What did we do? We noted $$\left(x+x^{-1}\right)^2-2=x^2+x^{-2}$$ and substituted $$u=x + \frac{1}{x}$$. The solutions for $$u^2+u=1$$ are $$-\varphi$$ and $$\varphi-1$$, $$\varphi$$ being the golden ratio. We now have two equations: $$x+\frac1x=-\varphi\implies x^2+\varphi x+1=0\implies x=\frac{\pm\sqrt{\varphi-3}-\varphi}{2}$$ $$x+\frac{1}{x}=\varphi-1\implies x^2+(1-\varphi)x+1=0\implies x=\frac{\pm\sqrt{-\varphi-2}+\varphi-1}{2}.$$

You can now manipulate it to get a $$a+bi$$ form. Alternatively, you can multiply $$(x-r)(x-r^{-1})$$, divide $$x^4+x^3+x^2+x+1$$ by it and discover which $$r$$ makes remainder zero. This would be somewhat long, so I won't explictly do it, but here is the idea.

• Now pull the same trick for the $n$th root :) – nbubis Dec 11 '13 at 22:38
• This trick, @Nathaniel, gets the real subfield of $\mathbb Q(\zeta_n)$, in other words, it gives you the minimal polynomial for $2\cos(2\pi/n)$. – Lubin Dec 22 '13 at 21:39
• @Ian, your ability/age is pretty high. How do you learn maths, care to share? – Pacerier Jan 30 '14 at 9:39
• @Pacerier Thanks for the kind words. I used to spend a lot of time simply jacking around here, Wikipedia and some libraries in my city. I also like history, so I try to be inspired by the giants of the past. This symmetry $r\to 1/r$ in particular I saw long ago in an old algebra book when I was studying for entering a military institution (I don't have interest on it anymore). Now, I am reading lecture notes for a national olympiad and a number theory book (Ireland and Rosen). I'll enter university soon, that will help me. – Ian Mateus Jan 30 '14 at 14:40
• @IanMateus, Do you have some good math books to recommend? – Pacerier Jan 30 '14 at 21:06

You pretty much have the right idea. We have:

$1^{1/5} = (e^{2\pi ki})^{1/5} = e^{2\pi k i/5}=\cos(2 \pi k/5) + i\sin(2 \pi k/5)$

And that's in $a+bi$ form. Letting $k = 0,\dots,4$ gives you all $5$ fifth roots of unity.