Exercise: Find the fifth and tenth roots of unity in algebraic form.

This is an early exercise in Ahlfors Complex Analysis.

What I have tried so far:

For the fifth roots I have tried reducing the problem to the fact that $\Re (1+z+z^2+z^3+z^4)=1+\cos\theta+\cos2\theta+\cos3\theta+\cos4\theta$ and $\Im (z^5-1)=\sin 5\theta$

Using identities for the above trig functions and converting to rectangular form I get $\Re (1+z+z^2+z^3+z^4)=8x^4+4x^3-6x^2-2x+1$ and $\Im (z^5-1)=16y^5-20y^3+5y$.

Factoring $8x^4+4x^3-6x^2-2x+1$ we get $(2x^2-1)(4x^2+2x-1)$. The factor $4x^2+2x-1$ provides me with all that I need The factor $2x^2-1$ is leading me believe that $\pm\frac{\sqrt{2}}{2}$ are real components of two of my fifth roots. This is not possible. By symmetry, I know I should only have 3 distinct real components (including $x=1$). Why should this "extraneous pair" $\pm\frac{\sqrt{2}}{2}$ show up?

Otherwise, the solutions for $4x^2+2x-1=0$ do agree with what I would expect. Also the 5 solutions for the imaginary component are agreeing with my expectations.

As far as finding the tenth roots of unity, I would like to avoid finding trig identities for these components. I am hoping that knowing the fifth roots can provide a shortcut.

I know there are other ways to approach this problem, but I am interested in salvaging something resembling this approach to the problem if possible.

  • $\begingroup$ You can write $z=e^{ix}$ $\endgroup$ – Archis Welankar Feb 14 '16 at 17:07
  • $\begingroup$ I know the polar form of the roots. This is not the problem. $\endgroup$ – Earl Feb 14 '16 at 18:28

By $\sin 5\theta=5\sin \theta-20\sin^{3} \theta+16\sin^{5} \theta$, put $\theta =\frac{\pi}{5}$ and $y=\sin \frac{\pi}{5}$, you have



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

Reject the upper case since $y^{2}<\sin^{2} \frac{\pi}{4}=\frac{1}{2}$

Can you proceed?

  • $\begingroup$ Yes. As I said, I have the imaginary solutions already. I would like to know why $\pm\frac{\sqrt{2}}{2}$ shows up as a solution to for the real component when they are clearly not valid real components for 5th roots. $\endgroup$ – Earl Feb 14 '16 at 19:01
  • $\begingroup$ $x^{2}=1-y^{2}=\frac{3+\sqrt{5}}{8}$, or $x=\frac{\sqrt{5}+1}{4}$ $\endgroup$ – Ng Chung Tak Feb 14 '16 at 19:04
  • $\begingroup$ For your reference mathworld.wolfram.com/TrigonometryAnglesPi5.html $\endgroup$ – Ng Chung Tak Feb 14 '16 at 19:10
  • $\begingroup$ I like that. But again, as I said in my post, I already have these solutions as well. I was hoping that someone could explain why the process I explain in my post seems to imply that $\pm\frac{\sqrt{2}}{2}$ is showing up as a solution. $\endgroup$ – Earl Feb 14 '16 at 19:11
  • 1
    $\begingroup$ Note that if $z^{4}=-1$, then $z^{4}+z^{3}+z^{2}+z+1=i(\pm 1\pm \sqrt{2})$ which is purely imaginary. So only equating real parts gives extra possibilities. $\endgroup$ – Ng Chung Tak Feb 14 '16 at 20:15

\begin{align*} \cos 5\theta &= \cos^5\theta + 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta \\ &= \cos^5\theta + 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-\sin^2\theta)^2 \\ &= ... \\ &= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta. \end{align*}


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