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Does there exist a function $f(z)$ holomorphic in $\mathbb{C}\backslash\{0\}$, such that

$$\left|f(z)\right|\geq\frac{1}{\sqrt{\left|z\right|}}$$

for all $z\in\mathbb{C}\backslash\{0\}?$

I'm not really sure on how to proceed or which particular theorems I should look at.

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    Have you considered $1/f$?2011-08-15
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    @Jones Meyer $1/f$ won't work, just check that for $z=4$ you have $1/4 = |f(z)| < \frac{1}{\sqrt{|z|}} = 1/2$.2011-08-15
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    @Thomas Klimpel: Considering $1/f$ is a good way to go, in my opinion. I don't understand what you wrote.2011-08-15
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    I don't think that such a function exists, it's just a question how to prove that. @Jonas Meyer Perhaps I misunderstood your suggestion. I thought you where proposing $f(z)=1/z$, so I pointed out that this doesn't work, as can be checked by inserting $z=4$ into the requested inequality.2011-08-15
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    @Thomas, why do you think Jonas was proposing $f(z)=1/z$, when what Jonas wrote was "Have you considered $1/f$"? In any event, have you considered $1/f$?2011-08-15
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    @Thomas: Thanks for clearing that up (I said $1/f$, not $1/z$). No, there is no such function, and I would approach it by contradiction, supposing such $f$ exists, and considering $g(z)=1/f(z)$ as the next step. I wonder if movesonthemove had any thoughts on this.2011-08-15
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    show that $1/f$ is holomorphic on $\mathbb{C}$, and then consider the proof of [Liouville's Theorem](http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28complex_analysis%29) using [Cauchy's Integral Formula](http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula).2011-08-15
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    @robjohn: Since $1/f(z)$ vanishes at 0, it’s of the form $zg(z)$ with $g$ entire.2011-08-15

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