Part B: The Basics of Linearity
It is given that the input $ e^{2jt} \! $ gives the output $ t*e^{-2jt} \! $ and that the input $ e^{-2jt} \! $ gives the output $ t*e^{2jt} \! $.
We are asked to give the system's response to an input of $ cos(2t) \! $. Recall that $ cos(2t)=\frac{1}{2}(e^{2jt}+e^{-2jt}) \! $. Note we are given the systems response to the inputs $ e^{2jt} \! $ and $ e^{-2jt} \! $. Thus, because the system is linear, we can easily see the system's output to this input will be $ \frac{t}{2}(e^{-2jt}+e^{2jt})=tcos(2t). $
