Re: [分析] 複變一題
※ 引述《thisday (小綱)》之銘言：
: If f(z) is analytic in |z|≦ 1 and satisfies |f| = 1 on |z| = 1
: show that f(z) is rational.
: 我有想用Blaschke product
Let a represent any root of f with order m in |z|<1.
Consider f(z)Π[(1-a*z)/(z-a)]^m := g(z)
a
where a* is the conjugate of a, then |g|=1 on |z|=1, and g has no root
in |z|<1, so by applying maximum modulus principle to g and 1/g we
get that g is a constant since |z|≦1 is compact, this shows that
f is rational.
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◆ From: 123.192.216.62
※ 編輯: bineapple 來自: 123.192.216.62 (12/17 05:01)
推
12/18 16:19, , 1F
12/18 16:19, 1F
※ 編輯: bineapple 來自: 123.192.216.62 (12/18 16:31)
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12/18 16:34, , 2F
12/18 16:34, 2F
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