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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.

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    $\begingroup$ Do you know the formula $e^{i\theta}=\cos \theta+i\sin \theta$? $\endgroup$
    – LASV
    Commented Dec 11, 2013 at 21:24

4 Answers 4

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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.

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    $\begingroup$ Now pull the same trick for the $n$th root :) $\endgroup$ Commented Dec 11, 2013 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
    Commented Dec 22, 2013 at 21:39
  • $\begingroup$ @Ian, your ability/age is pretty high. How do you learn maths, care to share? $\endgroup$
    – Pacerier
    Commented Jan 30, 2014 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
    Commented Jan 30, 2014 at 14:40
  • $\begingroup$ @IanMateus, Do you have some good math books to recommend? $\endgroup$
    – Pacerier
    Commented Jan 30, 2014 at 21:06
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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.

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Given the equation $z^5 = 1$ where $z \in \mathbb{C}$, subtract $1$ from both sides.

$0 = z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1)$

Now, for solving $z^4 + z^3 + z^2 + z + 1 = 0$, one can express it in the form $\text{square} = \text{square}$ by completing the square twice as follows:

$z^4 + z^3 + z^2 + z + 1 = 0$

First, add $z^2$ to both sides:

$\iff z^4 + z^3 + 2z^2 + z + 1 = z^2$

Next, group and factor out $z$ from $z^3 + z$.

$\iff [z^4 + 2z^2 + 1] + z(z^2 + 1) = z^2$

Note that $z^4 + 2z^2 + 1$ is a perfect square; it's equivalent to $(z^2 + 1)^2$.

$\iff (z^2 + 1)^2 + z(z^2 + 1) = z^2$

Complete the square by adding the square of half the 'coefficient' of $z^2 + 1$ to both sides.

$\iff (z^2 + 1)^2 + z(z^2 + 1) + \frac{z^2}{4} = z^2 + \frac{z^2}{4}$

$\iff (z^2 + \frac{1}{2}z + 1)^2 = \frac{5}{4}z^2$

Take the square root of both sides.

$\iff z^2 + \frac{1}{2}z + 1 = \frac{\pm_1\sqrt{5}}{2}z$

Subtract $\frac{\pm_1\sqrt{5}}{2}z$ from both sides of this new equation.

$\iff z^2 + \frac{1\mp_1\sqrt{5}}{2}z + 1 = 0$

But now, I can just follow the normal completing the square algorithm for solving quadratic equations, in order to determine the value of $z$. We have:

$z^2 + \frac{1\mp_1\sqrt{5}}{2}z = -1$

$\iff z^2 + \frac{1\mp_1\sqrt{5}}{2}z + \frac{6\mp_12\sqrt{5}}{16} = \frac{6\mp_12\sqrt{5}}{16} - 1$

$\iff (z + \frac{1\mp_1\sqrt{5}}{4})^2 = -\frac{10\pm_12\sqrt{5}}{16}$

$\iff z + \frac{1\mp_1\sqrt{5}}{4} = \frac{\pm_2\sqrt{10\pm_12\sqrt{5}}}{4}i$

$\iff z = \frac{-1\pm_1\sqrt{5}\pm_2\sqrt{10\pm_12\sqrt{5}}i}{4}$

Which was to be derived.

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You can also do this with algebra no trigonometry apart from the pythagorean theorem.

You know $z^5=1$ and $z=1$ is one of the solutions.

Factoring that out, you get $z^5-1=(z-1)(z^4+z^3+z^2+z+1)=0$

Suppose $z\ne 1$.

$z^4+z^3+z^2+z+1=0$

Equation remains true for the conjugates.

$\bar{z}^4+\bar{z}^3+\bar{z}^2+\bar{z}+1=0$

Subtract bottom from top and factor:

$(z-\bar{z})(z+\bar{z})(z^2+\bar{z}^2)+(z-\bar{z})(z^2+z\bar{z}+\bar{z}^2)+(z+\bar{z})(z-\bar{z})+(z-\bar{z})=0$

Divide by $(z-\bar{z})$ . We know this isn't zero since 1 is the only real root.

Note $z\bar{z}=1$ and $(z+\bar{z})^2=z^2+1+\bar{z}^2$. We can use this substitution to focus on the real part since if the real part is $x$ it satisfies $2x=z+\bar{z}$.

$(z+\bar{z})(z^2+\bar{z}^2+2-2) +(z^2+\bar{z^2}+2-1)+(z+\bar{z})+1=0$

$2x(4x^2-2)+(4x^2-1)+2x+1=0$

$8x^3+4x^2-2x=0$

$2x(4x^2+2x-1)=0 $

We know $x\ne 0$ since only $i$ and $-i$ fit that and they dont' sovle the main equation.

$4x^2+2x-1=0\implies \frac{-2\pm\sqrt{4+16}}{8}=\frac{-1\pm \sqrt{5}}{4}=x$

The real (x) and imaginary (y) parts satisfy $x^2+y^2=1$

$x^2= \frac{3 \mp \sqrt{5}}{8}$

$y^2=\frac{5 \pm \sqrt{5}}{8}$

Combine either of the real parts with either of the complex parts to get the 4 remaining roots of unity.

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