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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?

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

1 Answer 1

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

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