Set $V:=D_1(e^{\frac{i\pi}{6}})\cap D_1(e^{-\frac{i\pi}{6}})$. Find the (unique) explicit conformal function $f:V\rightarrow \mathbb{C}$ such that $f(V)=\mathbb{D}$ with $f(1)=0$ and $f'(1)>0$.
This result must be a consequence for my first problem here which say that any angular section $U=\{ z\mid-\alpha<Arg(z)<\alpha\}$ ($0<\alpha<\pi$) have a unique conformal function that send it to $\mathbb{D}$. If there exists another way please give me hint, I don't know how to attack this problem.
Remember that from the viewpoint of the Riemann sphere, circles and straight lines are the same.
The angular sector $U_\alpha = \{ z : -\alpha < \operatorname{Arg} z < \alpha\}$ is a domain bounded by two line segments (circular arcs on the sphere) intersecting in $0$ and $\infty$.
The domain $V = D_1(e^{\pi i/6}) \cap D_1(e^{-\pi i/6})$ is bounded by two circular arcs intersecting in $0$ and $\sqrt{3}$.
If you apply the Möbius transformation mapping $0\mapsto 0,\, 1 \mapsto 1,\, \sqrt{3}\mapsto \infty$ to $V$, what do you get?