Let $f(z)$ be analytic on $\mathbb{D}$ = {${z\in\mathbb{C}:|z-1|<1}$} such that $f(1) = 1$, if $f(z) = f(z^2)$ for all $z\in\mathbb{D}$,

Question

Let $f(z)$ be analytic on $\mathbb{D}$ = {${z\in\mathbb{C}:|z-1|<1}$} such that $f(1) = 1$, if $f(z) = f(z^2)$ for all $z\in\mathbb{D}$, then which of the following statements are correct?

1) $f(z) = [f(z)]^2$

2) $f(\frac{z}{2}) = \frac{1}{2}f(z)$

3) $f(z^3) = [f(z)]^3$

4) $f'(1) = 0$

I can't understand how to do solving this problem. please anyone give me some hints.

Zarrax · Answer 1 · 2014-06-08T15:11:12.733

4

Starting with $z_1 = {1 \over 2}$ and letting $z_{n+1} = \sqrt{z_n}$ (real square root) you obtain a sequence $z_1,z_2,...$ converging to $1$. By the relation $f(z) = f(z^2)$ each $f(z_n)$ has the same value. By continuity of $f(z)$ this value is $f(1) = 1$.

So you have a sequence of zeroes of $f(z) - 1$ converging to $z = 1$. Can you do the rest from here?

edited Jun 08 '14 at 15:11

answered Jun 08 '14 at 14:29

Zarrax

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score 3 · Answer 2 · answered Jun 08 '14 at 13:50

3

4) $f' (z) =2zf' (z^2 )$ hence $f' (1) =2f'(1)$ hence $f'(1)=0.$

answered Jun 08 '14 at 13:50

$1\in\mathbb{D}$ – Jun 08 '14 at 13:56
Yes we do.. $1\in\mathbb{D}$ on which $f$ is analytic! – FireGarden Jun 08 '14 at 13:58
Ah, I saw $\Bbb D$ and didn't read. Sorry. – Ted Shifrin Jun 08 '14 at 14:03

Let $f(z)$ be analytic on $\mathbb{D}$ = {${z\in\mathbb{C}:|z-1|<1}$} such that $f(1) = 1$, if $f(z) = f(z^2)$ for all $z\in\mathbb{D}$,

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