I'm trying to prove the result stated in the accepted answer of this SE post.
That is, let $f(z) = \frac{1}{z} \prod_{j=1}^n (z - a_j)^{\lambda_j}$, with the $a_j \in S^1$ distinct and oriented negatively (say), and the $\lambda_j > 0$ with $\sum_{j=1}^n \lambda_j = 2$. Now it turns out that given a 'star', $K = \bigcup_{j=1}^n [0,a_j]$ with $a_j \neq 0$ such that $\arg a_j \neq \arg a_k$ for $j \neq k$, there is a (essentially) unique conformal map from $\mathbb{D}$ to its exterior, mapping $0$ to $\infty$, precisely of the form $\frac{\alpha}{z} \prod_{j=1}^n (z - z_j)^{\frac{2p_{j-1}}{n}}$ where $\frac{2p_{j-1} \pi}{n}$ is the angle between the segments $[0,a_{j-1}]$ and $[0,a_j]$. Moreover all other such maps (to the exterior of the same star) are the same as this one except the $z_j$ are all rotated, i.e. we replace each one with some $u_j = Az_j$ with $|A| = 1$ and change $\alpha$ by a factor of absolute value one.
The post seems to claim that the converse is true, that is for any choice of $a_j$ on the unit circle and $\lambda_j > 0 $ summing to $2$, we get a conformal map from $\mathbb{D}$ to the exterior of some star.
It is clear to me that $f$ maps $\partial \mathbb{D}$ to a star with $n$ segments, but I do not see why we can assume such an $f$ is injective in the first place, i.e. why is it the case that $f$ cannot map a point in $\mathbb{D}$ to the star? Certainly $\partial f(\mathbb{D})$ is contained in the star, but I do not see why the other side of the inclusion is true.
Thanks in advance.
This is an answer combining peek-a-boo's previous answer with a small additional argument.
First of all, one can use the argument principle to show that for every $\zeta \in U$, where $U = \hat{\mathbb{C}} \setminus K$ and $K$ is the star with $n$ segments, there is precisely one point $z \in \mathbb{D}$ such that $f(z) = \zeta$.
We will now show that in fact $f$ is injective. We already know it assumes every value in $\hat{\mathbb{C}} \setminus K$ exactly once in $\mathbb{D}$, suppose it assumes some value in $K$ twice in $\mathbb{D}$, this value is necessarily nonzero (by the form of $f$, it has no zeros in $\mathbb{D}$), but in that case we will also assume some value in $\hat{\mathbb{C}} \setminus K$ at least twice in $\mathbb{D}$, because if we assume $f(p) \in K \setminus \{0 \}$ at $p$ and $q$, by the open mapping theorem disjoint neighbourhoods of $p$ and $q$ are mapped onto some ball centered at $f(p)$, and the preimage of some point in $\hat{\mathbb{C}} \setminus K$ contained in this small ball centered at $f(p)$ will contain points in both of these disjoint neighbourhoods, contradicting injectivity of $f$ on $f^{-1}(U)$.
But if $f$ is injective and $f(\partial \mathbb{D}) = K$, then certainly $f$ assumes no value of $K$ in $\mathbb{D}$, by essentially the same argument as in the paragraph above.
This shows that $f$ is a conformal mapping from $\mathbb{D}$ onto $\hat{\mathbb{C}} \setminus K$