若复数4-3i/3+ai为纯虚数,则（2+ai/2-ai）*2010=_______

4-3i/3+ai=(4-3i)(3-ai)/(3^2-(ai)^2)=(12-4ai-9i-3a)/(9+a^2)=(12-3a-(4a+8)i)/(9+a^2)
实部为0
12-3a=0
a=4
(2+4i)/(2-4i)=(2+4i)^2/(4+16)=(4+16i-16)/(20)=-12/20+16/20i=-3/5+4/5i
sina=4/5 cosa=-3/5
(cosa+isina)^2010=cos(2010a)+isin(2010a) 其中a=arcsin4/5

(4 - 3i)/(3 + ai) * (3 - ai)/(3 - ai)
= [(12 - 3a) + (- 4a - 9)i]/(a^2 + 9)
= (12 - 3a)/(a^2 + 9) + [(- 4a - 9)/(a^2 + 9)]i
纯虚数，即(12 - 3a)/(a^2 + 9) = 0 ==> a = 4
(2 + 4i)/(2 - 4i) * 2...