<s>[INST] How to show that this complex equation has 10 non real roots and how to express them 

I did the first part successfully:
$$w^{12}=1= \cos 2\pi + i \sin 2\pi$$
$$w= \cos \frac{\pi k}{6} + i \sin \frac{\pi k}{6}$$

Where $k=0,1,2,3,4,5,6,7,8,9,10,11$


I struggled with this for a long time, I know that the two real roots are $z=2$ and $z=-2$
But how do I prove it and show it can be expressed as that. Please  help.
Im new to these types  of questions

I only know how to do this when its something like: 
$$w^4=1$$
Where there is a the number $1$ or $-1$ and one sided equation. Because then I can write
$$w^4-1=0$$
$$(w-1)(w+1)(w^2+1)=0$$
 [/INST] By writing the second equation as
$$\Bigl(\frac{z+2}{z}\Bigr)^{12}=1$$
it reduces to the first.  Therefore the solutions are
$$\frac{z+2}{z}=\cos\frac{k\pi}{6}+i\sin\frac{k\pi}{6}\tag{$*$}$$
for $k=0$ or $k=6$ or $k=\pm1,\pm2,\ldots,\pm5$.  Now $k=0$ gives
$$\frac{z+2}{z}=1$$
which is impossible, and $k=6$ gives
$$\frac{z+2}{z}=-1$$
which has the real solution $z=-1$.  So this leaves ten non-real solutions.  You can find them from $(*)$ and do a bit of simplification using trig functions in order to arrive at the given answers.  Specifically, solve for $z$, rationalise the denominator and use half-angle formulae:
$$\eqalign{z
  &=-\frac{2}{1-(\cos\frac{k\pi}{6}+i\sin\frac{k\pi}{6})}\cr
  &=-2\frac{(1-\cos\frac{k\pi}{6})+i\sin\frac{k\pi}{6}}
    {(1-\cos\frac{k\pi}{6})^2+(\sin\frac{k\pi}{6})^2}\cr
  &=\cdots\cr
  &=-\frac{(1-\cos\frac{k\pi}{6})+i\sin\frac{k\pi}{6}}{1-\cos\frac{k\pi}{6}}
    \cr
  &=-1-i\frac{2\sin\frac{k\pi}{12}\cos\frac{k\pi}{12}}{2\sin^2\frac{k\pi}{12}}
    \ .\cr}$$
In fact, if you are familiar with complex exponentials you can write $(*)$ as
$$\frac{z+2}{z}=e^{k\pi i/6}$$
and the algebra becomes much easier.
Comment.  Worth a thought: including the real solution, we have $11$ solutions.  Can you explain why this is so, when your original equation involved polynomials of degree $12$?  Why are there not $12$ solutions?
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