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(Hi. This is my first question here.)

A well known result in complex analysis says that there is an $\varepsilon\gt 0$ such that if $f$ is holomorphic in (a neighborhood of) the closed disk ${\mathbb D}$ of radius 1, and $f'(0)=1$, then $f({\mathbb D})$ contains a disk of radius $\varepsilon$.

This is due to Landau and, accordingly, the largest possible $\varepsilon$ is called Landau's constant. The standard proof (see, for example, the book Complex Variables by Berenstein-Gay) gives $\varepsilon\ge1/16$.

As far as I understand, the best known bounds are $$ \frac 12\lt \varepsilon\le\frac{\Gamma(\frac13)\Gamma(\frac56)}{\Gamma(\frac16)}=0.54325\dots $$

However, I have been unable to locate any proofs of the first inequality, or any updated treatments of the second one (due to Rademacher). For all I know, current bounds may be better, or there may be a standard source to read about this.

Could you please give me some suggestions on where to look, or ideas on how to improve the $1/16$ bound, even if shorter of $1/2$?

(Many thanks!)

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2 Answers 2

up vote 3 down vote accepted

You can find some answers in Section 10.1 of Reinhold Remmert's Classical Topics in Complex Function Theory (Springer GTM 172) -- or in the original german edition Funktionentheorie 2.

A simple proof is given that Landau's constant $L$ verifies $L>\frac{3}{2}\sqrt{2}-2\simeq 0.1213$.

A more involved proof of $L>\sqrt{3}/4\simeq 0.4330$ is also presented: this is a consequence of Ahlfors's Theorem (Section 10.1.4 in the above reference, link).

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Many thanks! This certainly gets me started in the right direction. –  Bruce Dec 11 '10 at 7:06

Is the following reference helpful?

Current status of Bloch Constant and Landau Constant bounds

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Yes, it is! :-) (Embarrassing that I missed this question when I looked before posting.) Many thanks! –  Bruce Dec 11 '10 at 16:47
    
You are welcome! I find it independently, but this link was in the list of "related" links (on the right side of the screen). Such a list is automatically generated when you post a question. –  Petya Dec 11 '10 at 17:09

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