I'm having trouble solving the following exercise in complex analysis. Let $L_k:z=r\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n}}(1\leqslant r<\infty)$ be a ray for $k=1,\cdots,n$, and $\Omega=\mathbb C\setminus\bigcup\limits_{k=1}^nL_k$ the plane with $L_k$'s removed. We are to find a conformal mapping $\varphi$ from upper-half plane to $\Omega$, such that $\varphi(\mathrm{i})=0$.
My attempts and questions: I've tried to construct the conformal mapping by Schwarz-Christoffel formula, but I'm getting stuck on the domain $\Omega$.
- I guess that $\Omega$ can be regarded as a $2n$-polygon with $n$ vertices at infinity, where the angles at vertices $\mathrm{e}^{\mathrm{i}\frac{2k\pi}{n}}$ all equals to $2\pi$, and the angles at infinity vertices all equals to $-\frac{2\pi}{n}$. Is it right?
- How to choose suitable preimages of vertices of $\Omega$ on $\mathbb R\cup\{\infty\}$?
- Is there a way to solve this problem without using Schwarz-Christoffel formula?
p.s. The final answer given by the author is $$\varphi(z)=\frac{z^2+1}{\left(\sum_{k=0}^{[\frac n2]}(-1)^k\binom{n}{2k}z^{n-2k}\right)^{\frac 2n}}$$
Any help would be appreciated, thanks!