设方程x^2-(2-i)x+(2m-4)i=0的实数根为a
m=bi (a,b为实数)
则a^2-(2-i)a+(2bi-4)i=0,
根据复数相等的充要条件得
a^2-2a-2b=0且a-4=0
解得a=4,b=4
m=4i