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.

• 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)? – Scounged Jul 20 '15 at 15:22
• its the open disc. See the edition. – user149418 Jul 20 '15 at 15:24
• The third statement is not true, I believe, as you may take $f(z)=e^{2\pi i z}$ and $r=1$. – Anton Tselishchev Jul 20 '15 at 15:32

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 $f(z)=\sin(\frac{2\pi z}{r})$?
• For (1), under what conditions on $f$ will this hold? For (2), what have said, how to show that $f$ will then be constant? – user149418 Jul 20 '15 at 15:37
• @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. – Hagen von Eitzen Jul 20 '15 at 15:39
• If $f$ is an homeomorphism, then this certainly holds. – user149418 Jul 20 '15 at 15:42