2
$\begingroup$

I am stuck with the following problem: (GATE-Question)

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

I do not know how to progress with the problem or what property to use. Please help. Thanks in advance for your time.

2 Answers 2

5

(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 $f(0)=0$.

  • 0
    You are welcome.2012-12-14
2

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 $f(0)=0$.