Question about injective holomorphic functions on $\mathbb{D}$ and Koebe's quarter theorem Let $f$ be an injective holomorphic function on $\mathbb{D}$ such that $f(0) = 0$, $f'(0) = 1$. The open mapping theorem implies that $f(|z| < 1)$ contains an open neighborhood of the origin. Then $f(|z| < 1)$ contains a ball $\{|w| < r_{f}\}$ for some $r_{f} > 0$ depending on $f$. Is there a way to get rid of this dependence on $f$? I know that Koebe's one-quarter theorem says that we can take $r_{f} = 1/4$, but are there other ways to see this other than Koebe's quarter theorem?
 A: Yes, there are other ways, but they are more involved that the standard proof of the   $1/4$ theorem via Gronwall's area theorem. 
One way is to do what Koebe actually did; he only proved the existence of a uniform lower bound on $r_f$, without $1/4$. The "Koebe 1/4 theorem" as we know it is due to Bieberbach. The paper by Koebe is pretty long, and I gave up trying to figure out what he did. 
Instead, I'll present a normal family argument, based on Montel's theorem. Suppose the result is false; then there is a sequence $f_n$ of injective functions on $\mathbb D$ such that $f_n(0)=0$, $f_n'(0)=1$, and $f_n  $ omits $w_n$, where $w_n\to 0$. Let $$g_n(z) = \sqrt{1-w_n^{-1}f_n(z)} \tag{1}$$
where the branch is chosen so that $g_n(0)=1$. Note that (1) defines a holomorphic function in $\mathbb D$ because the content of square root is never $0$, and the domain is simply-connected. The function $g_n$ omits the values $0$ (as just said) and $-1$ (because $f_n(z)\ne 0$ for $z\ne 0$). By Montel's theorem, $\{g_n\}$ is a normal family. Since $g_n(0)=1$ for all $n$, the functions don't escape to infinity; so there must be a subsequence $g_{n_k}$ that converges to a holomorphic function $g$ uniformly on compact subsets of $\mathbb D$.  Then the derivatives of $g_{n_k}$ also converge uniformly on compact subsets. But this contradicts the fact that 
$$g_n'(0) = \frac1{2w_n}\to \infty \quad \text{as }\ n\to\infty $$

Note: the proof does not use the full assumption of injectivity; only the fact  that $f_n(z)\ne 0$ when $z\ne 0$ was used. I don't know what is the sharp lower bound for $r_f$ in this class.
