作者Bourbaki (大狐狸)
看板Math
标题[分析] 一题Stein复变上的习题
时间Tue May 17 20:03:29 2011
p.281 prob 3
Suppose Ω is a simply connected domain that excludes the three roots of the
polynomials 4z^3 - (g_2)z - (g_3). For ρ belonging Ω and ρ fixed, define
ω dz
I(ω)=∫ ------------------------------- ω belonging Ω
ρ [4z^3 - (g_2)z - (g_3)]^(1/2)
then I has an inverse given by p(z+α) for some α
that is I(p(z+α))=z for approprite α.
d
他的hint说证明----I(p(z+α)) = +1 or -1 再利用p是偶函数
dz
^^^^^^^^^^^^^^^^^^^^^^^^^这里有做到
但是我不知道怎麽用偶函数这个条件来让I(p(z+α))=z
请强者指教
谢谢^^
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◆ From: 140.112.25.15
※ 编辑: Bourbaki 来自: 140.112.25.15 (05/17 20:03)
1F：→ keroro321 :也许这个建议会对你有帮助,先从ρ附近来看. 05/19 06:43
2F：→ keroro321 :由Ω的定义,知道I(ω) is well-defined.!! 05/19 06:44
3F：→ keroro321 :ρ fixed,方程式p(z)=ρ has two roots (z_0,z_1). 05/19 06:45
4F：→ keroro321 :(I(p(z_i)))'=I'(p(z_i))*p'(z_i) (i=0,1) 05/19 06:45
5F：→ keroro321 :Becaus p is even (p' is odd), we can choose a 05/19 06:46
6F：→ keroro321 :root z_2 (z_0 or -z_0=z_1 (mod M)) such that 05/19 06:46
7F：→ keroro321 :(I(p(z_2)))'= 1, 从连续性知道 05/19 06:47
8F：→ keroro321 :存在 a nbd U of z_2 使得 p(U)包含於Ω 并且 05/19 06:48
9F：→ keroro321 :(I(p(z)))'= 1 , So I(p(z))= z+c(cons.)!! (z属於U) 05/19 06:49
10F：→ keroro321 :I(p(z_2))=I(ρ)= 0 =z_2+c , hence c=-z_2 05/19 06:49
11F：→ keroro321 :I(p(z+z_2)) = z , (z+z_2 属於U) 05/19 06:52
12F：→ keroro321 :所以在 ρ 附近做到了,之後应该可以从类似证明 05/19 06:53
13F：→ keroro321 :open connected in |R^n => path connected 05/19 06:53
14F：→ keroro321 :证到整个 Ω. 05/19 06:53