(Stein, Complex analysis, p.253) If $$F(z)= \int_1^z \frac{d\zeta}{(1-\zeta ^n)^{2/n}}$$ then $F$ maps
the unit disk conformally onto the interior of a regular polygon with
$n$ sides and perimeter
$$2^{\frac{n-2}{n}}\int_0^\pi(\sin\theta)^{-2/n}d\theta$$
I know about conformal maps from $\Bbb{H}$(upper half plane) to polygon given by Schwarz-Christoffel integral. So I tried to consider $F \circ G$ where $G: \Bbb H \to \Bbb D$ by $G(z)=\frac{i-z}{i+z}$. But change of variables does not make the integrand simple. $\zeta = G(w), d\zeta = -2idw/(i+w)^2$ so $d\zeta/(1-\zeta ^n)^{2/n}=cdw/((i+w)^n-(i-w)^n)^{2/n}$. So I cannot use Schwarz-Christoffel integral. What should I do? I tried also $\zeta=e^{i\theta}$ but don't know how to proceed.
Best Answer
Although this is not obvious, the Schwarz-Christoffel formula works the same for maps of the disk as it does for maps of halfplane. Explanation here.
The form of the derivative $\dfrac{d\zeta}{(1-\zeta ^n)^{2/n}}$ tells us that the map will create $n$ interior angles of size $\pi(1-2/n)$, which is exactly the interior angle of a regular $n$-gone. Of course, angles do not determine an $n$-gon for $n>3$. But we can also observe the symmetry. For this, it helps to change the lower limit of integration to $0$ (I don't understand why it's set at $1$ to begin with; in any case this only contributes a constant): $$\begin{split}F(e^{2\pi i/n }z) &= \int_0^{e^{2\pi i/n }z} \frac{d\zeta}{(1-\zeta ^n)^{2/n}} \\ & =\int_0^{z} \frac{e^{2\pi i/n }\,d\zeta}{(1-(e^{2\pi i/n }\zeta) ^n)^{2/n}} \\ & = e^{2\pi i/n } \int_0^{z} \frac{d\zeta}{(1-\zeta^n)^{2/n}} \\ & = e^{2\pi i/n } F(z)\end{split}$$ where the second step is a change of variable.
Thus, the image of $F$ has $n$-fold symmetry, which proves it is a regular $n$-gon.
Finally, the perimeter is computed by plugging $\zeta=e^{i\theta}$ into $$\int_{|\zeta|=1}\dfrac{d\zeta}{|1-\zeta^n|^{2/n}}$$ and simplifying a bit. Note that $|1-\zeta^n|^2=2-2\cos n\theta = 4\sin^2(n\theta/2)$ and use periodicity.