1. Calculate $\frac{(1+i)^{1989}}{(1-i)^{1989}-(1+i)^{1989}} \quad\left(i^{2}=-1\right)$.
Solution. Since

$$
(1+i)^{2}=1+2 i+i^{2}=1+2 i-1=2 i \text { and }(1-i)^{2}=1-2 i+i^{2}=1-2 i-1=-2 i,
$$

we have

$$
(1+i)^{1989}=(1+i)^{1989}(1+i)=(2 i)^{994}(1+i)=2^{994}\left(i^{2}\right)^{497}(1+i)=-2^{994}(1+i)
$$

and similarly, $(1-i)^{1989}=-2^{994}(1-i)$. Using these equalities, we get

$$
\frac{(1+i)^{1989}}{(1-i)^{1989}-(1+i)^{1989}}=\frac{-2^{994}(1+i)}{-2^{994}(1-i)+2^{994}(1+i)}=\frac{-(1+i)}{-(1-i)+(1+i)}=\frac{-(1+i)}{2 i}=\frac{-(1+i) i}{-2}=\frac{i-1}{2} .
$$