Characterizing all analytic map from some simply connected region $\Omega$ to $B(0,1)$ with prescribed values suppose $\Omega = \{re^{i\theta}:0<r<\infty, |\theta|<\pi/4\}$, 
(i) show that there exists an analytic mapping from $\Omega$ to $B(0,1)$ such that $g(1)=0, g(2)=1/2$
(ii)show that there does not exist an analytic mapping from $\Omega$ to $B(0,1)$ such that $g(1)=0, g(2)=3/4$
For the part(i), $z^2$ sends $\Omega$ to right half plane, then by $\frac{z-1}{z+1}$, we have the image as unit disk, then we have the desired map as $(\frac{z^2-1}{z^2+1})$, but this is not right. I am a bit confused how to characterize the analytic and conformal map from a simply connected region in $\Bbb C$ to $B(0,1)$ with prescribed values. In many textbook, I just learned how to construct a specific conformal map, any ideas will be appreciated. 
By the way, there is similar question as follows:
determine all analytic maps $f$ of $\{z: |z|<1\}$ into $\Bbb C - \{x: x \leq 0\}$ with $f(0)=-1, f(1/2)=-1/9$.
What is the general strategy to attack these problems? 
 A: Where did you find this problem? I am fairly confident that it is incorrect.
Instead, let me solve:

Suppose $\Omega = \{re^{i\theta}:0\leq r<\infty, |\theta|<\pi/4\}$, 
(i) show that there exists an analytic mapping from $\Omega$ to $B(0,1)$ such that $g(1)=0, g(2)=3/5$.
(ii) show that there does not exist an analytic mapping from $\Omega$ to $B(0,1)$ such that $g(1)=0, g(2)=3/4$.

(I have changed a “$<$” to a “$\leq$” within the definition of $\Omega$, and “$1/2$” to “$3/5$” within part (i).)
Your working for part (i) is mostly correct, except that you have composed the functions in the wrong order. If $g(z)$ is the map $z\mapsto z^2$ followed by $z \mapsto \frac{z-1}{z+1}$ then $g(z)= \frac{z^2-1}{z^2+1}$. This $g(z)$ is an analytic mapping from $\Omega$ to $B(0,1)$ such that $g(1)=0$ and $ g(2)=3/5$, as required.
Now suppose $\exists$ another analytic function $h(z)$ mapping $\Omega$ to $B(0,1)$ with $h(1)=0$. 
Then $f(z):= h(g^{-1}(z))$ is an analytic function mapping $B(0,1)$ to itself, with $f(0)=0$. 
Hence $f$ satisfies the conditions of the Schwartz lemma and we have that $f(z)=az$ for some $a \in \mathbb{C}$ with $\vert a \vert =1$
$$
\begin{align}
&\implies h(g^{-1}(z))=az \qquad &\forall z \in B(0,1)
\\
&\implies h(g^{-1}(g(z)))=ag(z) \qquad &\forall z \in \Omega
\\
&\implies h(z)=ag(z) \qquad &\forall z \in \Omega
\\
&\implies h(2)=ag(2)=\frac{3}{5}a
\end{align}
$$
Therefore any analytic function $h(z)$ that maps $\Omega$ to $B(0,1)$ with $h(1)=0$ must have that $h(2)\in \{z:\vert z \vert = \frac{3}{5} \} $. 
Since $\frac{3}{4} \notin \{z:\vert z \vert = \frac{3}{5} \}$, we have shown your part (ii) (and have also shown that the “$1/2$” in your original part (i) must be a mistake).
This method - creating a function from $B(0,1) $ to $B(0,1)$ and applying the Schwartz lemma - will hopefully be useful to you when solving similar problems!
A: I will post my answer here. The trick is just schwartz lemma;
For part (1)
suppose there is a map $h:D \to D$, where $D$ is the disk, 
then we construct a conformal map from $\Omega$ to $D$, that is $f=\frac{z^2-1}{z^2+1}$, but the task is to build  an analytic map, so we compose $h$ and $f$, then we have $g=h(\frac{z^2-1}{z^2+1})$, by the condition $g(1)=0, g(2)=1/2$, we have $h(0)=0, h(3/5)=1/2$, here we apply schwartz lemma, we find that such $h$ is reasonable and could exist since $h(3/5)=1/2<3/5$, therefore, we let $h(z)=5z/6$, by composing with $f$, we have the desired analytic map 
For part (2)
Still by Schwartz lemma, we find $h(3/5)=3/4>3/5$, which is a contradiction to Schwartz Lemma.  
