Question

i是虚数单位,（1+i/1-i）的2009次方=

写出过程，谢谢

Answer 1

(1+i)/(1-i)=(1+i)²/(1+i)(1-i)=2i/2=i
所以 [(1+i)/(1-i)]^2009=i^2009=i^2008·i=i

Answer 2

(1+i)/(1-i)=(1+i)^2/(1+1)=(1-1+2i)/2=i
故原式=i^2009=(i^4)^502*i=1*i=i