e的Z（Z为复数）次幂加上i等于零,求Z

Question

e的Z（Z为复数）次幂加上i等于零,求Z

Answer 1

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)