We can write \[(a + b)^{12} = b^{12} \left( 1 + \frac{a}{b} \right)^{12} = \left( 1 + \frac{a}{b} \right)^{12}.\]Note that $ \left( \frac{a}{b} \right)^{12} = \frac{a^{12}}{b^{12}} = 1, $ so $ \frac{a}{b} $ is also a 12th root of unity. Let $ \frac{a}{b} = e^{i \theta}, $ so $ 12 \theta $ is a multiple of $ 2 \pi, $ i.e $ \theta = \frac{k \pi}{6} $ for some integer $ k $. Then \begin{align*} (1 + e^{i \theta})^{12} &= (e^{i \theta/2} (e^{-i \theta/2} + e^{i \theta/2}))^{12} \\ &= e^{6 i \theta} (e^{-i \theta/2} + e^{i \theta/2})^{12} \\ &= e^{6 i \theta} \left( \cos \frac{\theta}{2} - i \sin \frac{\theta}{2} + \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right)^{12} \\ &= e^{6 i \theta} 2^{12} \cos^{12} \frac{\theta}{2} \\ &= 2^{12} e^{k \pi i} \cos^{12} \frac{k \pi}{12} \\ &= 2^{12} (\cos k \pi + i \sin k \pi) \cos^{12} \frac{k \pi}{12} \\ &= 2^{12} \cos k \pi \cos^{12} \frac{k \pi}{12}.\end{align*} We must find the number of different possible values of this expression over all integers $ k $. Note that $ \cos k \pi $ is always equal to 1 or $ -1, $ and $ \cos^{12} \frac{k \pi}{12} $ is a decreasing function for $ 0 \le k \le 6, $ giving us 7 different values. Furthermore, \[\cos k \pi = \cos (12 - k) \pi\]and \[\cos^{12} \frac{k \pi}{12} = \cos^{12} \frac{(12 - k) \pi}{12},\]so further values of $ k $ do not give us any new values of $ 2^{12} \cos k \pi \cos^{12} \frac{k \pi}{12} $. Hence, there are a total of $ \boxed{7} $ different possible values.