若复数z=(a-√2)+3i为纯虚数,a属于R,则(a+i^2009)/(ai-1)的值为多少?1楼计算错了_数学解答

若复数z=(a-√2)+3i为纯虚数,a属于R,则(a+i^2009)/(ai-1)的值为多少?
1楼计算错了

人气：431 ℃　时间：2020-04-11 18:46:08

解答

z=(a-√2)+3i为纯虚数则a-√2=0 a=√2
那么(a+i^2009)/(ai-1)
=(√2+i)/(√2i-1)
=(√2+i)(√2i+1)/(√2i-1)(√2i+1)
=(√2+i)(√2i+1)/(-3)
=(2i+√2-√2+i)/(-3)
=3i/(-3)=-i