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
Commented Dec 11, 2013 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 :) Commented Dec 11, 2013 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 $2\cos(2\pi/n)$. Commented Dec 22, 2013 at 21:39
• @Ian, your ability/age is pretty high. How do you learn maths, care to share? Commented Jan 30, 2014 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. Commented Jan 30, 2014 at 14:40
• @IanMateus, Do you have some good math books to recommend? Commented Jan 30, 2014 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.

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.

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.