Part B: The basics of linearity
$ e^{2jt} \rightarrow linear-system \rightarrow te^{-2jt} $
$ e^{-2jt} \rightarrow linear-system \rightarrow te^{2jt} $
From these two transformations we can tell that the system is as below:
$ x(t) \rightarrow linear-system \rightarrow y(t) = tx(-t) $
Therefore, when the input, x(t) = cos(2t), the output is as follows:
$ cost(2t) \rightarrow linear-system \rightarrow y(t) = tcos(-2t) $
