$ e^{2jt} $ yields $ te^{-2jt} $ and $ e^{-2jt} $ yields $ te^{2jt} $
and the system is linear
since Euler's formulat states that : $ e^{ix} = \cos x + i\sin x \! $
$ e^{2jt} $ yields $ te^{-2jt} $ and $ e^{-2jt} $ yields $ te^{2jt} $
and the system is linear
since Euler's formulat states that : $ e^{ix} = \cos x + i\sin x \! $