Existence of an injective holomorphic function with larger derivative Let $\Omega$ be a simply connected region with $0\in \Omega$, $\Omega \neq \mathbb{C}$. Suppose that $f$ is an one-to-one holomorphic function from $\Omega$ into the open unit disc $D$ such that $f(\Omega)\neq D$. Show that there is an one-to-one holomorphic function $g$ from $\Omega$ into $D$ such that $|f'(0)|<|g'(0)|$.
My attempt: Let $h$ be an automorphism of $D$. I want to prove that $g=h\circ f$ for some $h$. This equivalent to $|h'(f(0))|>1$ for some $h$, by chain rule. To deal with the problem, I let $h(z)=\dfrac{z-\alpha}{1-\bar{\alpha}z}$ for some $\alpha\in D$. However, I have no idea how I should find an appropriate $\alpha$. Does anyone have ideas?
 A: That is a standard argument used as part of the proof of the Riemann mapping theorem,
see for example THE RIEMANN MAPPING THEOREM:
W.l.o.g assume that $f(0)=0$, and $b \in \Bbb D \setminus f(\Omega)$.
$$
 T_b(z) = \frac{z-b}{1-\overline b z}
$$
is an automorphism of the unit disk. $(T \circ f)(\Omega) \subset \Bbb D$ is a
simply-connected domain not containing zero, therefore a holomorphic branch
$S$ of the square root exists on that domain. With $c = S(-b)$ we can now define
$$
 g = T_c \circ S \circ T_b \circ f
$$
which is an injective map from $\Omega$ into the unit disk with $g(0) = 0$.
It remains to show that $|f'(0)| < |g'(0)|$. With
$$
 \varphi(z) = (T_c \circ S \circ T_b)^{-1}(z) = T_b^{-1}((T_c^{-1}(z))^2)
$$
we have
$$
 f'(z) = \varphi'(g(z)) g'(z) \implies |f'(0)| = |\varphi'(0)| \cdot |g'(0)| \, .
$$
Finally, $|\varphi'(0)|< 1$ follows from the Schwarz Lemma because $\varphi$
maps the unit disk into itself, with $\varphi(0) = 0$, but is not a rotation.
A: Let $z_0=f(\Omega)$. $f$ is an one-to-one holomorphic function and $\Omega$ is simply connected region then $f(\Omega)$ is also simply connected region. By Riemann mapping theorem there exist an one-to-one holomorphic function $h:f(\Omega)\to D$ such that $h(z_0)=0$ and $h'(z_0)=2$. Set $g:\Omega\to D$ with $g=hof$ which is one-to-one holomorphic and 
$g'(0)=h'(z_0)f'(0)$ and $|g'(0)|>|f'(0)|$.
A: Edit @MartinR beat me to it with the same argument.
If $f(0)\neq 0$ then your approach will work. Now suppose $f(0)=0$ and $f$ avoids the value $a$ for some $0< a <1$. (This condition on $a$ is just for convenience. It can always be met by considering $\omega f$ for some $\lvert \omega \rvert=1$ instead.) Let $$h(z) = \left(\frac{a - f(z)}{1 - a \ f(z)}\right)^{\frac12}.$$ This function is single valued on $\Omega$, injective, maps into $D$, and $h(0) = \sqrt{a}$. Finally let $$g(z) = \frac{\sqrt{a} - h(z)}{1 - \sqrt{a}\  h(z)}.$$ Then $$g’(0) = \frac{1+a}{2 \sqrt{a}}\  f’(0).$$ Now note that $$\frac{1+a}{2 \sqrt{a}} = \frac{(1-\sqrt{a})^2 + 2 \sqrt{a}}{2 \sqrt{a}}> 1.$$
