<s>[INST] Show that the element $z=i \cos \frac{\pi}{3}+\sin \frac{\pi}{3} = i( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3})$ belongs to $U_{12}$ Show that the element $\displaystyle z=i \cos \frac{\pi}{3}+\sin \frac{\pi}{3} = i( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3})$ belongs to U12
What I don't understand:


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*In what way $i \cos \frac{\pi}{3} + \sin \frac{\pi}{3}$ is different from $i(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3})$ corresponding to the polar solutions of U12


2.U12 has 12 polar elements and from Cayley's table we see there are 4 elements, how this is different?
The 4 elements are the generators and others are not (primes), how's this is different for the polar elements?
Thanks for the help and tips
 [/INST] I might be misunderstanding your question, but here goes nothing! 
It seems like we want to prove that $z^{12} = 1$. Now
$$ z = i(\cos\frac{\pi}{3} - i\sin\frac{\pi}{3}) = i(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = ie^{-i\frac{\pi}{3}} = ie^{-i\frac{2\pi}{6}} $$ 
$$z^6 = i^6(e^{-i\frac{2\pi}{6}})^6 = -1$$
$$z^{12} = (z^6)^2 = (-1)^2 = 1$$
For your second question, notice that $z$ is the $12$-th root of $1$. If another element $\alpha$ satisfies $\alpha^{12} = 1$, see if you can figure out how $z$ relates to $\alpha$
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