# $f:\mathbb C\rightarrow \mathbb C$ be an arbitrary analytic function satisfying $f(0)=0$ and $f(1)=2.$

I am stuck with the following problem:

Let $f:\mathbb C\rightarrow \mathbb C$ be an arbitrary analytic function satisfying $f(0)=0$ and $f(1)=2.$ Then, which of the following items is correct?

(a) there exists a sequence $\{z_{n}\}$ such that $|z_{n}|$ and $|f(z_{n})|> n$,
(b) there exists a sequence $\{z_{n}\}$ such that $|z_{n}|$ and $|f(z_{n})|< n$,
(c) there exists a bounded sequence $\{z_{n}\}$ such that $|z_{n}|$ and $|f(z_{n})|> n$,
(d) there exists a sequence $\{z_{n}\}$ such that $z_{n} \rightarrow 0$ and $f(z_{n})\rightarrow 2.$

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(a) Yes, by Liouville theorem (as we have an entire function which is not constant).

(b) Take $z_n=0$ for all $n$.

(c) Remember that $f$ is continuous, so it maps bounded sets to bounded ones.

(d) No, as $f$ is continuous at $0$ and $f(0)=0$.

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How would (a) follow from Liouville theorem? –  Sugata Adhya Dec 14 '12 at 14:55
$f$ being non-constant entire $\exists$ {$z_n$} such that $|f(z_n)|>n$. This part is easy. But how would $|z_n|>n$? –  Sugata Adhya Dec 14 '12 at 15:01
The sequence is necessarily unbounded. Extract a subsequence if necessary. –  Davide Giraudo Dec 14 '12 at 15:05
Now I get it. Thanks. –  Sugata Adhya Dec 14 '12 at 15:06
You are welcome. –  Davide Giraudo Dec 14 '12 at 15:22

Hints:

a) You know that $f$ is non-constant. So, (BLANK)'s theorem says that $f$ must be un(BLANK).

b) Using the fact that every holomorphic function is (BLANK) the fact that $f(0)=0$ let's you choose such a sequence.

c) Using the fact that every holomorphic function is (BLANK) again, and using the fact that every bounded sequence sits inside a (BLANK) set you know that the values of the sequence must be bounded.

d) This, once again, can't happen because every holomorphic function is (BLANK) and $f(0)=0$.

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