Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Would you correct me? Which of the following statements are true?

a. There exists an entire function $f : \mathbb{C}\rightarrow\mathbb{C}$ which takes only real values. and is such that $f(0) = 0$ and $f(1) = 1$

b. There exists an entire function $f : \mathbb{C}\rightarrow\mathbb{C}$ such that $f(n + 1/n) = 0$ for all $n\in\mathbb{N}$

c. There exists an entire function $f : \mathbb{C}\rightarrow\mathbb{C}$ which is onto and which is such that $f(1/n) = 0$ for all $n\in\mathbb{N}$

Well, for a) it is only constants.

b) can have such non-constants entire function.

c) only constants.

Are my answers correct?plz help.

share|cite|improve this question
They are all correct if you remove the word "onto" from c (constant functions don't satisfy that). – Matt Apr 26 '12 at 21:34
For (a) if $c$ is a constant, then how can you have $c=0$ and $c=1$? Similarly for (c). Ironically, (b) is the only one for which there is a constant solution for $f$. (But I doubt that all solutions are constant) – Hurkyl Apr 26 '12 at 21:34
up vote 3 down vote accepted

a. Not true because of the open mapping theorem. No nonconstant function can take on only real values, regardless of the other condition in the problem. Note that a constant function $f$ cannot satisfy $f(0)=0$ and $f(1)=1$, so we have to rule out constant solutions too, for trivial reasons.

b. You can construct a function with zeroes only at these points using the Weierstrass Factorization Theorem.

c. The zeros of $f$ have an accumulation point, so such an $f$ must be 0 everywhere.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.