- Other examples of elliptic integrals providing conformal maps from the upper half-plane to rectangles are given below.
- The function
\[ \int_ {0} ^ {z} \frac {d \zeta}{\sqrt {\zeta (\zeta - 1) (\zeta - \lambda)}}, \quad \mathrm{with} \lambda \in \mathbb {R} \mathrm{and} \lambda \neq 1 \]
maps the upper half-plane conformally to a rectangle, one of whose vertices is the image of the point at infinity.
- In the case \(\lambda = - 1\) , the image of
\[ \int_ {0} ^ {z} \frac {d \zeta}{\sqrt {\zeta (\zeta^ {2} - 1)}} \]
is a square whose side lengths are \(\textstyle { \frac { \Gamma ^ { 2 } ( 1 / 4 ) } { 2 { \sqrt { 2 \pi } } } }\)
Solution
For part (a), the finite branch points must be distinct, so the intended hypothesis is \(\lambda\in\mathbb R\setminus\{0,1\}\). Choose the branch of \[ F'(z)=\frac1{\sqrt{z(z-1)(z-\lambda)}} \] that is holomorphic on \(\mathbb H\). The Schwarz–Christoffel formula says that a factor \((z-a)^{-\beta}\) at a real prevertex produces interior angle \((1-\beta)\pi\). At each of \(0,1,\lambda\) the exponent is \(-1/2\), so each corresponding angle is \(\pi/2\).
At infinity put \(w=1/z\). Since \(F'(z)=O(z^{-3/2})\), \[ \frac{d}{dw}F(1/w)=-w^{-2}F'(1/w)=O(w^{-1/2}), \] so infinity is a fourth prevertex with angle \(\pi/2\). Hence the image is a quadrilateral with four right angles, therefore a rectangle.
For \(\lambda=-1\), \[ F'(z)=\frac1{\sqrt{z(z^2-1)}}. \] The symmetry \(z\mapsto -z\) interchanges the two finite intervals adjacent to \(0\), and \(z\mapsto1/z\) interchanges the corresponding opposite sides; hence the rectangle has equal adjacent side lengths and is a square.
One side has length \[ L=\int_0^1\frac{dx}{\sqrt{x(1-x^2)}}. \] With \(t=x^2\), \[ L=\frac12\int_0^1 t^{-3/4}(1-t)^{-1/2}\,dt =\frac12 B\left(\frac14,\frac12\right) =\frac{\Gamma(1/4)\sqrt\pi}{2\Gamma(3/4)}. \] Euler’s reflection formula at \(1/4\) gives \[ \Gamma(1/4)\Gamma(3/4)=\pi\sqrt2, \] and therefore \[ L=\frac{\Gamma(1/4)^2}{2\sqrt{2\pi}}. \]