解方程急需答案!!!

2024-11-29 21:03:13
推荐回答(3个)
回答1:

回答2:

设z= x+yi, x, y 是R
e^z = e^x * e^yi = e^x *(cosy + i siny)
所以
e^x *(cosy + i siny) = 1 +√3 i
e^x *cosy = 1
e^x *sin y = √3
两式相除
tan y = √3
y = π/3 +2kπ

e^x siny = √3/2 * e^x = √3
e^x = 2
x = ln2

z = ln2 + (π/3 +2kπ) i
k 是z

回答3:

e=a+bi
则a²-b²+2abi=1+√3i
a²-b²=1
2ab=√3
b=√3/(2a)

所以a²-3/(4a²)=1
4a⁴-4a²-3=0
(2a²-3)(2a²+1)=0
a²=3/2
b²=a²-1=1/2
所以
e=√6/2+i√2/2,e=-√6/2+-i√2/2