Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be an entire function such that $\lvert f\rvert$ approaches infinity as $\lvert z\rvert$ tends to infinity. Then

  1. $f(1/z)$ has an essential singularity at $0$.
  2. $f$ cannot be a polynomial.
  3. $f$ has finitely many zeros.
  4. $f(1/z)$ has a pole at $0$.

Option 2 is false, what about 1, 3, 4?

share|improve this question
    
This appears to be a duplicate of this question. –  robjohn May 24 '13 at 11:07
add comment

1 Answer

(1). Hint: Consider $f(w) := w$. Has $z \mapsto f \left( \frac{1}{z} \right)$ an essential singularity at 0?

(3). Since $|f(w)| \to \infty$ as $|w| \to \infty$ there exists $R>0$ such that $|f(w)| \geq 1$ for all $|w| \geq R$ (which implies that there are no zeros in $\{w \in \mathbb{C}; |w| \geq R\}$). Assume that there exist countable many zeros in $B[0,R]$. Since $B[0,R]$ is compact there exists a convergent subsequence, i.e. there exists a sequence $(w_k)_{k \in \mathbb{N}}$ in $B[0,R]$ such that $f(w_k) =0$, $w_k \to w$. Now apply the identity theorem (see here, section "An Improvement").

(4). There exists a sequence $(a_n)_n$ such that $f(w) = \sum_{n \in \mathbb{N}_0} a_n \cdot w^n$ for all $w \in \mathbb{C}$. Since $|f| \to \infty$ as $|w| \to \infty$ there exists $k>0$ such that $a_k \not= 0$. So what can you say about $$f \left( \frac{1}{z} \right) = \sum_{n \in \mathbb{N}_0} a_n \cdot z^{-n}?$$

(notation: $B[0,R] := \{w \in \mathbb{C}; |w| \leq R\}$)

share|improve this answer
add comment

Your Answer

 
discard

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.