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.
Best Answer
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
$0=5y-20y^{3}+16y^{5}$
$0=5-20y^{2}+16y^{4}$
$\displaystyle y^{2}=\frac{5 \pm \sqrt{5}}{8}$
Reject the upper case since $y^{2}<\sin^{2} \frac{\pi}{4}=\frac{1}{2}$
Can you proceed?