- If
\[ F (z) = \int_ {1} ^ {z} \frac {d \zeta}{(1 - \zeta^ {n}) ^ {2 / n}}, \]
then \(F\) maps the unit disc conformally onto the interior of a regular polygon with n sides and perimeter
\[ 2 ^ {\frac {n - 2}{n}} \int_ {0} ^ {\pi} (\sin \theta) ^ {- 2 / n} d \theta . \]
Solution
Factor \[ 1-z^n=\prod_{k=0}^{n-1}(1-\omega^k z), \qquad \omega=e^{2\pi i/n}. \] Thus \[ F'(z)=(1-z^n)^{-2/n} \] has Schwarz–Christoffel exponent \(-2/n\) at each of the \(n\) equally spaced boundary points \(\omega^k\). Hence every image vertex has interior angle \[ \left(1-\frac2n\right)\pi=\frac{n-2}{n}\pi. \] Moreover \(F'(\omega z)=F'(z)\), and therefore \[ F(\omega z)-F(0)=\omega\bigl(F(z)-F(0)\bigr). \] So the image polygon is invariant under rotation by \(2\pi/n\). A polygon with \(n\) equal angles and this rotational symmetry is a regular \(n\)-gon.
It remains to compute its perimeter. On the boundary arc from \(1\) to \(e^{2\pi i/n}\), write \(z=e^{i\theta}\), \(0\le\theta\le2\pi/n\). Its image is one side, of length \[ \begin{aligned} L &=\int_0^{2\pi/n}|1-e^{in\theta}|^{-2/n}\,d\theta\\ &=2^{-2/n}\int_0^{2\pi/n}\sin(n\theta/2)^{-2/n}\,d\theta\\ &=\frac{2^{1-2/n}}{n}\int_0^\pi(\sin u)^{-2/n}\,du. \end{aligned} \] Multiplying by \(n\) gives the perimeter \[ 2^{1-2/n}\int_0^\pi(\sin u)^{-2/n}\,du =2^{(n-2)/n}\int_0^\pi(\sin u)^{-2/n}\,du. \]