4
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Do you know the formula $e^{i\theta}=\cos \theta+i\sin \theta$? $\endgroup$ – LASV Dec 11 '13 at 21:24
12
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ Now pull the same trick for the $n$th root :) $\endgroup$ – nbubis Dec 11 '13 at 22:38
  • $\begingroup$ This trick, @Nathaniel, gets the real subfield of $\mathbb Q(\zeta_n)$, in other words, it gives you the minimal polynomial for $2\cos(2\pi/n)$. $\endgroup$ – Lubin Dec 22 '13 at 21:39
  • $\begingroup$ @Ian, your ability/age is pretty high. How do you learn maths, care to share? $\endgroup$ – Pacerier Jan 30 '14 at 9:39
  • $\begingroup$ @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. $\endgroup$ – Ian Mateus Jan 30 '14 at 14:40
  • $\begingroup$ @IanMateus, Do you have some good math books to recommend? $\endgroup$ – Pacerier Jan 30 '14 at 21:06
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.