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I have the following question:let $f(z)$ be continuous on the closed unit disk, $\{z: |z|\leq 1\}$, and analytic on the open unit disk, $\{z: |z|<1 \}$, with $f(e^{it})=0$ for $0\leq t \leq \pi/4$. Prove that $f(z)=0$ on the open unit disk.

I try to apply maximum modules principle but I couldnt get the result. Any help would be great.

Jonas Meyer
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user135582
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1 Answers1

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Consider $g(z)= \prod_{k=0}^7 f(e^{ki\pi/4}z).$

zhw.
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  • Is the exponent of e $k \pi i /4$, I mean there is an i in the exponent? So this new function g will have absolute value on the boundary of the disk that implies the absolute value of the function inside the disk is also zero hence g is zero on the open unit disk which than f is also zero on the open unit disk? – user135582 Apr 07 '15 at 02:52
  • Thanks, yes I forgot the $i$, will put it in now. – zhw. Apr 07 '15 at 03:27
  • So did I interpret your hint correctly? Thanks a lot. – user135582 Apr 07 '15 at 03:31