f(z)＝(az+b)/(cz+d)
0,1+i,i 通过分式线性变换 0,2,无限...
f(0)=0
f(1+i)=2
f(i)=∞
f(0)＝(b)/(d)＝0
b=0
f(i)=∞
f(i)＝(ai)/(ci+d)=∞
ci+d＝0
f(1+i)=2
f(1+i)＝(a(1+i)+b)/(c(1+i)-ci)
=(a(1+i))/(c)=2
a/c=2/(1+i)
f(z)＝(az+b)/(cz+d)
＝(az)/(cz-ci)
=(2z)/[(1+i)(z-i)]
=[(z)(1-i)]/(z-i)
f(z)＝[(z)(1-i)]/(z-i)
不动点
z=[(z)(1-i)]/(z-i)
z＝0或z=1
[f(z)-0]/[f(z)-1]={[(z)(1-i)]/(z-i)-0}/{[(z)(1-i)]/(z-i)-1}
=[(z)(1-i)]/{[(z)(1-i)]-z+i}
=[(z)(1-i)]/[-(iz)+i]
=[(z)(1-i)]/(z-1)(-i)
=[1-i][(z-0)/(z-1)]
f^n(z)=Q(z)后
n次迭代
[f(z)-0]/[f(z)-1]=a(1,z)
[f(f(z))-0]/[f(f(z))-1]=a(2,z)
[f(f(f(z)))-0]/[f(f(f(z)))-1]=a(3,z)
a(n,z)=[1-i]^n[(z-0)/(z-1)]
[Q(z)-0]/[Q(z)-1]=[1-i]^n[(z-0)/(z-1)]
1/[Q(z)-1]=[1-i]^n[(z-0)/(z-1)]-1
[Q(z)-1]=1/{[1-i]^n[(z-0)/(z-1)]-1}
Q(z)=1/{[1-i]^n[(z-0)/(z-1)]-1}+1
f^n(z)=1/{[1-i]^n[(z-0)/(z-1)]-1}+1
表达式求出来了,我不知道什么是恒等啊!你告诉我什么是恒等,往下我就会做啦!
如果f^n(z)=z
1/{[1-i]^n[(z-0)/(z-1)]-1}+1 ＝z
[1-i]^n[(z-0)/(z-1)]-1＝1/(z－1)
[1-i]^n[(z-0)/(z-1)]＝(z-0)/(z－1)
[(1-i)^n-1][(z-0)/(z-1)]=0
n>=1时
|(1-i)^n-1|>0
所以
(1-i)^n-1不＝0
(z-0)/(z-1)＝0
f^n(z)=z如果是恒等变换
要求对复数域内全体（除了z＝i）都有f^n(z)=z
有无数根,矛盾
所以不是恒等变换
