If $Ω$ is a simply connected proper sub-domain of $\mathbb{C}$, then show that there is a one-one bounded analytic function on $Ω$. If $Ω$ is a simply connected proper sub-domain of $\mathbb{C}$, then show that
there is a one-one bounded analytic function on $Ω$ without and using riemann mapping theorem

can anyone help me please to solve the problem
 A: Potato points out that since $\Omega$ a proper subset, you can translate so that $0$ is in the complement of $\Omega$. Since $\Omega$ is 1-connected, you can choose a path from $0$ to $\infty$. Take a branch of the square root on the complement of the path. This maps to a half-plane. Now conformally map the half-plane to the unit disk, scale if necessary, and translate if necessary to a subset of $\Omega$. All functions involved are injective holomorphic, hence their composition is injective holomorphic.
My original answer is below.
If $\Omega$ is bounded, restrict the identity map to $\Omega$.
If $\Omega$ is not dense in $\mathbb{C}$, translate so that the complement of $\Omega$ has $\mathbb{C}$ as an interior point. Now invert. The result is bounded. Scale if necessary. Now translate back to a subset of $\Omega$.
If $\Omega$ is dense in $\mathbb{C}$, translate so that $0$ is in the complement of $\Omega$. By simple-connectedness, we can connect $0$ to $\infty$ by a single line in the complement. Choose this line as a branch cut for the square root. The image of this is not dense in $\mathbb{C}$, so apply the operations of the previous case.
In each of these cases, we have used analytic injective functions (translation, inversion, dilation, branch of square root), so their composition is analytic and injective.
A: I believe there is an issue with the other answer, so here's the solution given in Ahlfors' text. There is some point $a\notin \Omega$. Since the region is simply connected, we can define a branch of $\sqrt{z-a}$. Call this $h(z)$. This in injective and does not take "opposite values": $w$ and $-w$ cannot both be in the range. Fix a point $z_0\in \Omega$. Then the image $h(\Omega)$ covers a disc of radius $r$ at $h(z_0)$ for some $r>0$, so by the opposite value remark, it does not meet the disc of radius $r$ centered at $-h(z_0)$.
You can now compose the above with a translation of $-h(z_0)$ to $0$ and then take $1/z$. The resulting region is bounded because there is an open neighborhood around $0$ before you take the reciprocal, so there will be an open neighborhood around infinity after.
