$12 \cdot 2$ If $i^{2}=-1$, then $(1+i)^{20}-(1-i)^{20}$ equals
(A) -1024 .
(B) $-1024 i$.
(C) 0 .
(D) 1024 .
(E) $1024 i$.
(25th American High School Mathematics Examination, 1974)
[Solution] Since $i^{2}=-1$, we have $(1+i)^{2}=2 i$ and $(1-i)^{2}=-2 i$.
And $(1+i)^{20}-(1-i)^{20}=\left[(1+i)^{2}\right]^{10}-\left[(1-i)^{2}\right]^{10}$, thus $\quad(1+i)^{20}-(1-i)^{20}=(2 i)^{10}-(-2 i)^{10}=0$. Therefore, the answer is $(C)$.