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  1. Let $D=B(z_0,R)$ be the open disc centered at $z_0$ with radius $R>0$ and $f$ be a non-constant entire function. Is it true that $f$ maps the boundary $\partial D$ of $D$ into the boundary $\partial f(D)$ of $f(D)$?

  2. Let $r>0$ and $f$ be an entire function such that $f(z+\frac{r}{n})=f(z)$ for all $z\in\Bbb C$ and all positive integer $n$. Is $f$ constant?

  3. Let $r>0$ and $f$ be an entire function satisfying $f(z+r)=f(z)$ for all $z\in\Bbb C$. Is $f$ constant?

For (1), I think Maximum Modulus theorem and Open Mapping theorem are to be applied. I have managed to show that interior goes to the interior. But I did not able to show that boundary goes to boundary, I think $f$ needs to be one-one to make it hold. Is that right?

For (2), the given condition implies that for any $z\in \Bbb C$, $f'(z)=0$ (using the definition and taking $h=r/n\to 0$). Hence $f$ is constant. I think this is ok.

For $(3)$, I don't know how to approach. I have tried many but those don't lead to any conclusion.

Any help is appreciated.

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  • $\begingroup$ Is $D=B(z_0,R)$ a circle in the complex plane with the midpoint $z_0\in \mathbb{C}$ and the radius $R$ in (1)? $\endgroup$ – Scounged Jul 20 '15 at 15:22
  • $\begingroup$ its the open disc. See the edition. $\endgroup$ – user149418 Jul 20 '15 at 15:24
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    $\begingroup$ The third statement is not true, I believe, as you may take $f(z)=e^{2\pi i z}$ and $r=1$. $\endgroup$ – Anton Tselishchev Jul 20 '15 at 15:32
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  1. Consider $f(z)=z^3$, $z_0=1, R=1$). Then $0\in \partial D$, and $0\in\partial f(D)$. But in $\partial f(D)$, $0$ is isolated; hence for boundary points sufficiently close to $0$, you obtain a counterexample.

  2. You are right. Actually, it would already be sufficient if thie condition holds for one $z\in\mathbb C$.

  3. What about $f(z)=\sin(\frac{2\pi z}{r})$?

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  • $\begingroup$ For (1), under what conditions on $f$ will this hold? For (2), what have said, how to show that $f$ will then be constant? $\endgroup$ – user149418 Jul 20 '15 at 15:37
  • $\begingroup$ @user149418 The idea in 1 is that $f$ "wraps over", so (as you somewhat suspected) if $f$ fails to be one-to-one, there is usually a boundary point thet is mapped to the interior of the image. $\endgroup$ – Hagen von Eitzen Jul 20 '15 at 15:39
  • $\begingroup$ If $f$ is an homeomorphism, then this certainly holds. $\endgroup$ – user149418 Jul 20 '15 at 15:42

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