Part B: The basics of linearity
$ e^{2jt} \rightarrow linear-system \rightarrow te^{-2jt} $
$ e^{-2jt} \rightarrow linear-system \rightarrow te^{2jt} $
The input, cos(2t) is equal to $ \frac{1}{2}(e^{j2t} + e^{-j2t}) $
From the properties of a linear system $ ax_1(t) + bx_2(t) \rightarrow linear system \rightarrow ay_1(t) + by_2(t) $
