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Let $f$ be an entire function such that $\;\lim\limits_{z\to\infty} |f(z)| =\infty\;$ .Then $f$ has at least one zero in $\;\Bbb C\;$ . How to prove this ? I am stuck. Please help.

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    Hint: what can you say about $\frac{1}{f(z)}$? – 2017-01-19
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    Rob's hint is excellent. I think you meant $\;|z|\to\infty\;$ in your limit... – 2017-01-19
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    it is 0 as limit – 2017-01-19
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    Will you please describe it in details sir? – 2017-01-19
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    @DonAntonio what's the difference? If anything the change should be $= + \infty$ instead of $= \infty$, but it's fine as is. – 2017-01-19
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    Do you know "Liouville's Theorem"? – 2017-01-19
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    Yes but trying to link with it – 2017-01-19
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    Here are the relevant facts about $\frac{1}{f(z)}$. (1) Either it's entire or $f(z)$ has a zero. (2) it's bounded (because, for all sufficiently large $z$, $|f(z)| > 1$, giving $f(z) \le 1$). Now what do you know about bounded entire functions? – 2017-01-19
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    Is it 1/z????or 1/f(z)??? – 2017-01-19
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    $\frac{1}{f(z)}$ - spotted in time for me to fix the comment. – 2017-01-19
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    @quid The difference is the agreemnet , or lack of, in understanding with a complex variable what $\;z\to\infty\;$ means. If, for example, the real part is kept fixed, say equal to one, and the imaginary part tends to zero, is it ok to write $\;z\to\infty\;$ ? I think it is not, and thus $\;|z|\to\infty\;$ makes it cyrstal clear that *both variables* (i.e., the real and imaginary pats) tend to infinity. Now, **if it is agreed** that this is the meaning of $\;z\to\infty\;$ then there's no problem, indeed... – 2017-01-19
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    @DonAntonio the meaning of $z \to \infty$ is the same as that of $z \to z_0$ for any $z_0$ in the topological space "one-point compactifcation of the complex plane." This may not have been taught exactly like this but something equivalent to it ought to appear when Liouville etc is discussed. – 2017-01-19

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Suppose f has no zero. Then $\frac{1}{f(z)}$ is a well defined holomorphic function. Since $|f(z)| \to \infty$ as $z \to \infty$, we know that $ \frac{1}{|f(z)|} \to 0$. So lets say for all $|z| > R$ we have that $\frac{1}{|f(z)|} \leq 1$. Also, the closure of $B_R(0)$ is compact and therefore $\frac{1}{f(z)}$ is bounded there as well. We now have an entire, bounded function. Hence it has to be constant by Liouvilles theorem. So your function is either constant or has a zero.

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    But it cannot be constant as lim is infinity. – 2017-01-19
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    @mathislove This started with "Suppose that $f$ has no zero". Then it was shown that "$f$ has no zero" $\implies$ $f = 0$, or is constant. We know that isn't possible (because of the infinite limit, as you mentioned), so it must be true that $f$ has at least one zero. This is a proof by contradiction. – 2017-01-19
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    But I am getting ,1/f is constant and nothing else.how to get f=0 from that? – 2017-01-20
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    If it has no zero, 1/f is constant and therefore f is constant and not the zero-function. There is no complex number s.t. 1/c=0. And if it is constant, it can not fulfil the limit condition. Its a proof by contradiction. – 2017-01-20