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?

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