Find explicitly a conformal mapping $f$ of $R$ onto the unit disc $U$ such that $f(1)=0$ and $f'(1)\gt 0$ Question
Let $R$ be the complex plane with the non positive real axis taken out. 
Find explicitly a conformal mapping $f$ of $R$ onto the unit disc $U$ such that $f(1)=0$ and $f'(1)\gt 0$ 

solution:
We know that all automorphism of the unit disk are of the form: 
$$f(z)=e^{i\theta}\frac{z-\alpha}{1-\overline{\alpha}}$$ for $|\alpha|<1$
Let $g(z)$ be a conformal map. And let's define a conformal map $$f(z)=e^{i\theta}(\frac{g(z)-g(z_0)}{1-\overline{g(z_0)}g(z)})$$
Then $f(1)=0 \iff g(1)=g(z_0)$
And $$f'(z)=\frac{g'(1).e^{i\theta}}{1-|g(1)|^2}$$
From here, I dont know how to reach its result? Is this true way to solve this question? If it is true, how to continue? Thanks. 

A hint:
The map  $\psi:\ z\mapsto {\rm pv}(z^{1/2})$ ("taking the positive square root") maps $R$ conformally onto the right half-plane $H$, and $\psi(1)=1$.
Now find a map $\phi$ from $H$ to $U$. By symmetry you can arrange $\phi(1)=0$.
