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}) $
$ 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}) $