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

complex-analysis

(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 $f(0)=0$.


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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