设z=a+bi,则 　　e^z=e^(a+bi)=e^a*e^bi=e^a*(cosb+isinb)(欧拉公式e^ix=cosx+isinx) 　　e^z+i=0=>e^acosb+ie^asinb+i=0 　　=>e^acosb+(e^asinb+1)i=0 　　由复数性质知,e^acosb=0,e^asinb+1=0 　　e^a≠0,∴cosb=0,sinb=-e^(-a)a=-ln(-sinb)=-ln[-sin(2kπ-π/2)] 　　∴z=a+bi=-ln[-sin(2kπ-π/2)]+(2kπ-π/2)i,(k∈整数)