- We consider conformal mappings to triangles.
- Show that
\[ \int_ {0} ^ {z} z ^ {- \beta_ {1}} (1 - z) ^ {- \beta_ {2}} d z, \]
with \(0 < \beta _ { 1 } < 1 , \ 0 < \beta _ { 2 } < 1\) , and \(1 < \beta _ { 1 } + \beta _ { 2 } < 2\) , maps \(\mathbb { H }\) to a triangle whose vertices are the images of 0, 1, and \(\infty ;\) , and with angles \(\alpha _ { 1 } \pi\) , \(\alpha_2\pi\), and \(\alpha _ { 3 } \pi _ { \ i }\) where \(\alpha _ { j } + \beta _ { j } = 1\) and \(\beta _ { 1 } + \beta _ { 2 } + \beta _ { 3 } = 2\)
What happens when \(\beta _ { 1 } + \beta _ { 2 } = 1 ?\)
What happens when \(0 < \beta _ { 1 } + \beta _ { 2 } < 1 ?\)
In (a), the length of the side of the triangle opposite angle \(\alpha _ { j } \pi\) is \(\begin{array} { r l } { { \frac { \sin ( \alpha _ { j } \pi ) } { \pi } \Gamma ( \alpha _ { 1 } ) \Gamma ( \alpha _ { 2 } ) \Gamma ( \alpha _ { 3 } ) } } \end{array}\)
Solution
Let \[ F(z)=\int_0^z \zeta^{-\beta_1}(1-\zeta)^{-\beta_2}\,d\zeta \] with branches chosen holomorphically on \(\mathbb H\). At the finite prevertices \(0\) and \(1\), the Schwarz–Christoffel exponents are \(-\beta_1\) and \(-\beta_2\), hence the interior angles are \[ \alpha_1\pi=(1-\beta_1)\pi, \qquad \alpha_2\pi=(1-\beta_2)\pi. \] Near infinity, \[ F'(z)\sim C z^{-(\beta_1+\beta_2)}. \] Under \(w=1/z\) this becomes \[ \frac d{dw}F(1/w)\sim C' w^{\beta_1+\beta_2-2}=C'w^{-\beta_3}, \qquad \beta_3=2-\beta_1-\beta_2. \] Thus the angle at the image of infinity is \[ \alpha_3\pi=(1-\beta_3)\pi=(\beta_1+\beta_2-1)\pi. \] When \(1<\beta_1+\beta_2<2\), all three \(\alpha_j\) lie in \((0,1)\) and sum to \(1\), so the image is a triangle with precisely these angles.
If \(\beta_1+\beta_2=1\), then \(\alpha_3=0\). The two sides issuing toward infinity become parallel and the image is an infinite strip-type degenerate triangle; analytically, \(F(z)=C\log z+O(1)\) at infinity.
If \(0<\beta_1+\beta_2<1\), then \(F(z)\sim C z^{1-\beta_1-\beta_2}\) at infinity, so the two boundary rays separate by the angle \((1-\beta_1-\beta_2)\pi\). The image is an unbounded polygonal region with the finite angles \((1-\beta_1)\pi\) and \((1-\beta_2)\pi\) and an exterior vertex at infinity rather than a bounded triangle.
For part (d), the side opposite \(\alpha_3\pi\) is the image of \([0,1]\), so its length is \[ L_3=\int_0^1 x^{-\beta_1}(1-x)^{-\beta_2}\,dx =B(\alpha_1,\alpha_2) =\frac{\Gamma(\alpha_1)\Gamma(\alpha_2)}{\Gamma(\alpha_1+\alpha_2)}. \] Since \(\alpha_1+\alpha_2=1-\alpha_3\), Euler reflection gives \[ \frac1{\Gamma(1-\alpha_3)} =\frac{\sin(\pi\alpha_3)}{\pi}\Gamma(\alpha_3), \] so \[ L_3=\frac{\sin(\pi\alpha_3)}{\pi} \Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3). \] The same argument after permuting the three prevertices gives, for each \(j\), \[ L_j=\frac{\sin(\pi\alpha_j)}{\pi} \Gamma(\alpha_1)\Gamma(\alpha_2)\Gamma(\alpha_3). \]