# Conformal mapping

Let $F(z)=\int_0^z z^{-\beta_1} (1-z)^{-\beta_2} dz$ with $0<\beta_1, \beta_2<1$ and $1<\beta_1 + \beta_2 <2$. Prove that $F$ maps the upper half-plane to a triangle whose vertices are the images of 0,1, and $\infty$ with angles $\alpha_1\pi , \alpha_2\pi, \alpha_3\pi$ where $\alpha_i +\beta_i =1$ and $\beta_1 +\beta_2 +\beta_3=2$ and the length of the side of triangle opposite angle $\alpha_i \pi$ is $\frac{sin(\alpha_i \pi)}{\pi} \Gamma(\alpha_1) \Gamma(\alpha_2) \Gamma(\alpha_3)$
What happen if $\beta_1 +\beta_2=1$? If $0<\beta_1 + \beta_2 <1$?