Re: [分析] 有請複變高手來幫忙~

[keroro321 (日夕)]

※ 引述《represent1 (represent)》之銘言：
: 一題複變：
: Suppose that f is an entire 1-1 function.  show that f(z)= az+b
: 有請複變高手來幫忙~ 感恩!

Choose a point z0 so that f'(z0)≠0.
Let
       g(z)=f(z+z0)－f(z0)

then   g(z) is an entire ,1-1 function. ( g(0)=0 , g'(0)≠0 )

    ∵ g(0) = 0 ,

    ∴ g(z) = z * h(z)    where h(z)≠0 for all z .

為了獲得關於 h(z) 在"∞"附近函數的行為,所以觀察 g(z) 在"∞"附近時的表現.

       g'(0)≠0 , there exists open neighborhoods U of 0 , V of 0 such that
                                        ╴╴
       g(U) = V .       (suppose V 包含 Br(0) (半徑r>0閉球),U 取 bounded)

    ∵ g : 1-1 function =>  ||g(z)|| > r   for all z 屬於 C－U .

                               1         1
                    (   =>  ──── < ──   )
                            ||g(z)||     r

                    1
       Let s(z) = ───  , then z=0 is a removable singularity of s(z) .
                  g(1/z)

       Take s(0) = lim s(z) = a = 0  (if a≠0 => g(z) bounded => g(z)=c →←)
                   z→0

       Again s(0) = 0 ,

          so s(z) = (z^n) * t(z)   where t(0)≠0 , n:some positive integer

              s(z)               z
       lim  ─── = lim    ───────  = t(0) ≠ 0
       z→0   z^n    z→0   (z^n) * h(1/z)


       Either   h(z)   or   1/h(z)   is bounded !
               ───       ───
               ( n=1)       (n >1)

       so g(z) = z * h(z) = z * c
       ￣￣￣￣￣￣￣￣￣￣￣￣￣ #




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