Is there an example of a schlicht function $f(z)=z+a_2z^2+a_3z^3+\cdots$, which is analytic and injective on the open unit disk $\mathbb{D}$, such that $1/a_2$ belongs to the range $f(\mathbb{D})$? Or is $1/a_2$ necessarily an omitted value of $f$?
It is easy to design a function like $z/(1az)$ with small $a$ that maps the circle to a nice domain whose closure does not contain $1/{a_2}$. Now take this domain and grow a blob that contains $1/{a_2}$ deep in its interior but is connected to our initial domain by a very thin stem. Then the conformal mapping to this new domain has almost the same coefficients as the original one (almost being controlled by the width of the stem, not the size of the blob), so after renormalizing you get a counterexample.

1$\begingroup$ Is it possible to be more explicit? I don't see why the small perturbation on $a_2$ induced by the stem and blob can be guaranteed not to send $1/a_2$ outside the blobsince the perturbation on $1/a_2$ may be large. $\endgroup$– P GibsonMar 22 '12 at 0:59

1$\begingroup$ The perturbation in $a_2$ is not large at all. When the stem width tends to $0$, the conformal mapping tends to the mapping to the original domain uniformly on compact subsets of the unit disk. Despite the blob is large, its full preimage is just a tiny piece near the very boundary of the circle, which doesn't affect the behavior near the center. That is one of rather counterintuitive properties of the conformal mappings. If you want, I can post a full proof but you can just as well read some decent complex analysis textbook. $\endgroup$– fedjaMar 22 '12 at 1:31

1$\begingroup$ OK, I see nowletting the stem width go to zero produces a convergent sequence whose limit is forced to be z/(1az) itself. (So the blob is handled progressively later in the tails of the respective series.) Thanks for the answer. $\endgroup$– P GibsonMar 22 '12 at 16:05