Question

一道复变函数题，求具体解析

一道复变函数题，求具体解析第二题怎么求，求具体解析

Answer 1

z=x+yi
(z-1)/(z+1)
=[(x-1) +yi] /[ (x+1)+yi ]
=[(x-1) +yi] .[ (x+1)-yi ] /[ (x+1)^2 +y^2 ]
=[ (x^2-y^2-1) +2yi]/[ (x+1)^2 +y^2 ]
实部= (x^2-y^2-1) /[ (x+1)^2 +y^2 ]
虚部= 2yi /[ (x+1)^2 +y^2 ]

追问

第二题。

根号3减i的m次方等于1+i的n次方，求最小正整数m,n

追答

(√3-i)^m =(1+i)^n
w=√3-i
|w|=2 , argw = -π/3
w=2[cos(-π/3) +isin(-π/3)]
w^m =2^m. [cos(-mπ/3) +isin(-mπ/3)]
z=1+i
|z| =√2, argz=π/4
z= √2[cos(π/4) +isin(π/4)]
z^n =(2)^(n/2) [cos(nπ/4) +isin(nπ/4)]
w^m = z^n
=>2^m. [cos(-mπ/3) +isin(-mπ/3)] =(2)^(n/2) [cos(nπ/4) +isin(nπ/4)]
2^m. cos(-mπ/3) =(2)^(n/2) .cos(nπ/4) (1) and
2^m. sin(-mπ/3) =(2)^(n/2) . sin(nπ/4) (2) and
(2)/(1)
tan(-mπ/3 ) = tan(nπ/4)
kπ -mπ/3 =nπ/4
(3k-m)π/3 = nπ/4
4(3k-m) = 3n
4m+3n = 12k
k=2, m=3, n=4
最小正整数 : m=3, n=4

追问

但是化成指数形式的话，不是存在一个2的m次方=根号2的n次方，不就是2m=n么，就不对了啊

追答

不
实属部分=实属部分

虚数部分=虚数部分
e.g
3(x+yi) = 4(a+bi)
=>
3x=3a
3y =4b

追问

对了，就是两复数的模相应的乘方相等嘛

追答

2^m. [cos(-mπ/3) +isin(-mπ/3)] =(2)^(n/2) [cos(nπ/4) +isin(nπ/4)]

[2^m. cos(-mπ/3) -(2)^(n/2).cos(nπ/4)] +i[2^m.sin(-mπ/3)] -(2)^(n/2) .sin(nπ/4)]] =0
=>
2^m. cos(-mπ/3) =(2)^(n/2).cos(nπ/4) and
2^m.sin(-mπ/3)] =(2)^(n/2) .sin(nπ/4)