Solution
= Solution
Since $(1+i)/(1-i)=i$ and $(1+i)^2=2i$,
$$
\frac{(1+i)^{2019}}{(1-i)^{2017}}
=i^{2017}(1+i)^2
=i(2i)=\boxed{-2}.
$$
= Solution
Since $(1+i)/(1-i)=i$ and $(1+i)^2=2i$,
$$
\frac{(1+i)^{2019}}{(1-i)^{2017}}
=i^{2017}(1+i)^2
=i(2i)=\boxed{-2}.
$$