(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!)



Landau's constant
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](http://books.google.com/books?id=BHc2b0iCoy8C&lpg=PA230&ots=9hNpeOlrgu&dq=ahlfors%2520theorem%2520schlicht&pg=PA230#v=onepage&q=ahlfors%2520theorem%2520schlicht&f=false)).


