Cho số phức z thỏa mãn $z={{\left( \dfrac{1+i}{1-i}...

Ta có: $\dfrac{1+i}{1-i}=\dfrac{\left( 1+i \right)\left( 1+i \right)}{1+1}=\dfrac{2i}{2}=i\Rightarrow z={{\left( \dfrac{1+i}{1-i} \right)}^{2019}}={{i}^{2019}}\Rightarrow {{z}^{4}}={{\left( {{i}^{2019}} \right)}^{4}}={{\left( {{i}^{4}} \right)}^{2019}}=1$.