first,
$ e^{2jt}=cos(2t) + jsin(2t) $
$ e^{-2jt}=cos(2t) - jsin(2t) $
then we put into system, we got..
$ e^{2jt}\rightarrow system\rightarrow t*e^{-2jt} $
$ e^{-2jt}\rightarrow system\rightarrow t*e^{2jt} $
it means
$ e^{2jt} = t*(cos(2t)-jsin(2t)) $
$ e^{-2jt} = t*(cos(2t)+jsin(2t)) $
using Euler's formula
$ e^{ix} = cosx + isinx $
input is
$ x(t)=\cos(2t) $
output will be..
$ \frac{1}{2}*tcos(2t) + \frac{1}{2}tcos(2t) = tcos(2t) $
so,
$ cos(2t)\rightarrow system \rightarrow tcos(2) $
