1. **State the problem:** Find the 12th term of the sequence: $1+i$, $2i$, $-2+2i$, ... 2. **Identify the pattern:** The given sequence terms are complex numbers: $a_1=1+i$, $a_2=2i$, $a_3=-2+2i$. 3. **Find the difference between consecutive terms:** $$a_2 - a_1 = 2i - (1+i) = -1+i$$ $$a_3 - a_2 = (-2+2i) - 2i = -2$$ The differences are not constant, so this is not an arithmetic sequence. 4. **Check the ratio between terms:** $$\frac{a_2}{a_1} = \frac{2i}{1+i} = \frac{2i(1 - i)}{(1 + i)(1 - i)} = \frac{2i - 2i^2}{1 - i^2} = \frac{2i - 2(-1)}{1 - (-1)} = \frac{2i + 2}{2} = 1 + i$$ $$\frac{a_3}{a_2} = \frac{-2+2i}{2i} = \frac{-2}{2i} + \frac{2i}{2i} = -\frac{1}{i} + 1 = i + 1$$ Since both ratios equal $1 + i$, the sequence is geometric with common ratio $r = 1 + i$. 5. **Find the 12th term using geometric sequence formula:** $$a_n = a_1 r^{n-1}$$ So, $$a_{12} = (1+i)(1+i)^{11} = (1+i)^{12}$$ 6. **Simplify $(1+i)^{12}$:** Recall that $1 + i = \sqrt{2} e^{i\pi/4}$ by Euler's formula. Therefore, $$ (1+i)^{12} = (\sqrt{2})^{12} \left(e^{i\pi/4}\right)^{12} = (\sqrt{2})^{12} e^{i3\pi} $$ Since $(\sqrt{2})^{12} = (2^{1/2})^{12} = 2^{6} = 64$ and $e^{i3\pi} = \cos 3\pi + i \sin 3\pi = -1 + 0i$, we get $$ (1+i)^{12} = 64 (-1) = -64 $$ 7. **Final answer:** The 12th term of the sequence is $\boxed{-64}$.