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Let $f(z)$ be analytic function on $D = \{z\in C : |z-1|<1\}$ such that $f(1) = 1$. If $f(z) = f(z^2)$ for all $z\in D$. Then which of the following statement is not correct.

1-$f(z) = [f(z)]^2$ for all $z\in D$

2- $f(z/2)$ = $\frac{1}{2}[f(z)]$ for all $z\in D$

3- $f(z) = [f(z)]^3$ for all $z\in D$

4- $f'(1) = 0$

I am fully stucked on this problem. I need help. Thanks

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1 Answer 1

up vote 3 down vote accepted

I claim that the hypotheses on $f$ imply that $f\equiv 1$. Indeed, note that $$f(2^{-1}) = f(2^{-1/2}) = f(2^{-1/4}) = \cdots = f(2^{-1/2^n}) = \cdots.$$ The sequence $2^{-1/2^n}$ converges to $1$ as $n\to \infty$, and thus $f$ takes the same value on a sequence with a limit point in $D$. It follows that $f$ must be constant. Since $f(1) = 1$ by hypothesis, $f\equiv 1$.

Since $f\equiv 1$, option (2) is wrong.

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Dear sir one option must be wrong. –  srijan May 12 '12 at 14:46
    
@srijan: Are you sure you wrote down (1)-(4) correctly? –  froggie May 12 '12 at 14:47
    
Oh i am extremely sorry sir. You were right . I edited option 2. –  srijan May 12 '12 at 14:51
    
@srijan: No problem, edited my answer to reflect that. –  froggie May 12 '12 at 14:52
    
How can you say that $f$ is identically one? Can you please explain. I would be really thankful to you. –  srijan May 12 '12 at 14:54
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