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which of the following statements are true?
1.an entire function $f$ such that $f(z) \neq 0$ in $\mathbb{C}$ and $\lim _{z \to \infty } f(z) \neq 0$ is necessarily constant.
2. if $f=u+iv$ is entire and $u^2 \leq v^2 +2004$ on $\mathbb{C}$ then $f$ is necessarily constant.


my thought:-

  1. it is true. if $f(z)$ is bounded then by liouvilles theorem we have done. otherwise if its not bounded then then $1/f$ is analytic and bounded. hence constant.
  2. I have no idea.

can anyone help me please.

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  • $\begingroup$ 1. Since there are functions for which neither $f$ nor $1/f$ are bounded, you need to take care with how you are using the hypotheses when you prove "if its not bounded then $1/f$ is analytic and bounded." 2. $u$ and $v$ are harmonic. Does this help? $\endgroup$ – Matt Apr 4 '13 at 2:28
  • $\begingroup$ then how I solve 1 $\endgroup$ – poton Apr 4 '13 at 2:38
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    $\begingroup$ 1. Consider $f(z)=e^z$. $\endgroup$ – i707107 Apr 4 '13 at 2:44
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For the first question, we consider two cases:

  • $\lim_{z \to \infty} f(z) = \infty$ implies that $f$ is a polynomial. But a polynomial with no roots must be constant.

  • $\lim_{z \to \infty} f(z) = L \in \Bbb{C}$ implies that $f$ is bounded. To see this, choose $R > 0$ such that $|z| > R \implies |f(z) - L| < 1$. Then $|f(z)| < L + 1$ for all $|z| > R$, and $\{z : |z| \le R\}$ is compact and $f$ is bounded on this set as well.

The second question can be answered positively via Picard's Theorem: $f$ can never take on values such as $1000 + 0i$ or $1000 + 1i$. Since $f$ misses two values, it must be constant.

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    $\begingroup$ Your answer to the second one is nice; but for the first one, you seem to be assuming that the limit $\lim_{z \to \infty} f(z)$ exists — how do you know this? $\endgroup$ – Peter LeFanu Lumsdaine Sep 8 '13 at 0:43
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    $\begingroup$ @PeterLeFanuLumsdaine It's implied in the statement of (1) that the limit exists. If the limit does not exist, it is false: $f(z) = e^z$ is the usual counterexample. $\endgroup$ – user61527 Sep 8 '13 at 0:45
  • $\begingroup$ Ah — so you’re reading “$\lim_{z \to \infty} f(z) \neq 0$” as meaning “the limit exists and is not equal to 0”? I would have read it as “it’s not the case that $\lim_{z \to \infty} f(z) = 0$”. I guess it’s a slightly ambiguous phrasing; but anyhow, you’ve now given answers for both possible readings of it! $\endgroup$ – Peter LeFanu Lumsdaine Sep 8 '13 at 0:48
  • $\begingroup$ @PeterLeFanuLumsdaine Yes, that's how I'm reading it. I agree that it's ambiguous, though. $\endgroup$ – user61527 Sep 8 '13 at 0:49

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