Problem: If i2=−1i^{2}=-1i2=−1, then (1+i)20−(1−i)20(1+i)^{20}-(1-i)^{20}(1+i)20−(1−i)20 equals
Answer Choices:
A. −1024-1024−1024
B. −1024i-1024i−1024i
C. 000
D. 102410241024
E. 1024i1024i1024i Solution:
Since i2=−1,(1+i)2=2ii^{2}=-1,(1+i)^{2}=2 ii2=−1,(1+i)2=2i and (1−i)2=2i(1-i)^{2}=2 i(1−i)2=2i. Writing (1+i)20−(1−i)20=((1+i)2)10−((1−i)2)10(1+i)^{20}-(1-i)^{20}=\left((1+i)^{2}\right)^{10}-\left((1-i)^{2}\right)^{10}(1+i)20−(1−i)20=((1+i)2)10−((1−i)2)10. we have (1+i)20−(1−1)20=(2i)10−(−2i)10=0(1+i)^{20}-(1-1)^{20}=(2 i)^{10}-(-2 i)^{10}=0(1+i)20−(1−1)20=(2i)10−(−2i)10=0.