The task was whether $\mathbb{C}^{*}$ can be conformally mapped to $\mathbb{C} \backslash \pm 1$
Then I read that: https://matmor.unam.mx/~robert/cur/2010-1%20CA/CA6.pdf
$\mathbb{C}^{*}$ is conformally equivalent to $\mathbb{C} \backslash$ {p} for any p ∈ $\mathbb{C}$, but not to any other subset of $\mathbb{C}$
I may not understand this correctly, but does it mean that $\mathbb{C}^{*}$ can be only mapped to $\mathbb{C} \backslash$ $\{p\}$, when $p$ is a single value? so I can not exclude 2 values (as in $\pm 1$) at the same time?
My questions: How can I show that:
$\mathbb{C}^{*}$ is conformally equivalent to $\mathbb{C} \backslash$ {p} for any p ∈ $\mathbb{C}$, but not to any other subset of $\mathbb{C}$
and how would the maps look like?
The map $z\mapsto z+p$ does the job for $\mathbb C^*$ to $\mathbb C/\{p\}$.
The topology of $\mathbb C^*$ and $\mathbb C/\{\pm \}$ are different. They are not even homeomorphic to each other, and definitely not holomorphic.
To show $\mathbb C^*$ is not conformally equivalent to any other subset of $\mathbb C$. Say $f:C^*\xrightarrow{\sim} X\subset\mathbb C$, since $\mathbb C^*$ is not homeomorphic to $\mathbb C$, $X$ misses at least one point of $\mathbb C$. By shifting, without loss of generality, we may assume $X$ misses $0$ as well.
Suppose $f$ is a biholomorphic map from $\mathbb C^*$ to its image. Then since $f$ is injective, by Great Picard, $0$ can only be either removable or a pole. If $0$ is removable, then $f$ is entire, and thus by Picard's little, the analytic extension of $f$ can miss at most one point, and hence $f(\mathbb C^*)$ can miss at most two, but we have already shown it cannot miss two, due to topology. If $0$ is a pole, then we have $g(z):=\frac{1}{f(1/z)}$ is a biholomorphic map from $\mathbb C^*$ to $X'$ where $X'$ is the image of $X$ under the bihomolorphic map $z\mapsto 1/z$ on $\mathbb C^*$ that contains $X$. Since $0$ is removable for $g$, we conclude as above $g(\mathbb C^*)$ misses only one point, and hence $f(\mathbb C^*)$ only misses one point as well.
If we more carefully track the details in the above argument, it can be shown that the biholomorphic group of $\mathbb C^*$ is generated by $z\mapsto az, z\mapsto 1/z$ for nonzero $a$, hence the group is isomorphic to $\mathbb C^*\rtimes (\mathbb Z/2\mathbb Z)$