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When proofing the Riemann mapping theorem that every simply connected domain $\mathbb{C} \neq D \subset \mathbb{C}$ is conformal equivalent to the unit disk $\mathbb{D}$, we can construct a Möbius transformation $g$ that maps $D$ to a bounded simply connected domain $T \subset \mathbb{D}$. Then we can construct a conformal mapping $h$ that maps $T$ onto $\mathbb{D}$.

In the textbook I'm reading they use a rotation $\theta : \mathbb{D} \rightarrow \mathbb{D}$ around $0$ afterwards and say that the function $\theta \circ h \circ g$ is the seeked conformal map from $D$ onto $\mathbb{D}$.

Why do we need the rotation $\theta$ afterwards? Is the function $h \circ g$ not already the conformal map we are looking for?

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  • $\begingroup$ I cannot see any reason for using $\theta$. In fact, $h \circ g$ is conformal. $\endgroup$ – Paul Frost Oct 23 '18 at 15:39
  • $\begingroup$ I found out now. The rotation is not necessarily needed to construct a conformal map. If you extend the riemann mapping theorem and demand the uniqueness of the conformal map you will have to use a rotation so you get: $f'(z_0) > 0$ $\endgroup$ – Arjihad Oct 23 '18 at 18:54
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The riemann mapping theorem is as follows

Is $\mathbb{C} \neq D \subset \mathbb{C}$ simply connected, then $D$ is conformally equivalent to the open unit disk $\mathbb{D}$.

You can extend the theorem by the following:

$z_0 \in D$. There is a unique conformal mapping $f:D \rightarrow \mathbb{D}$ with $f(z_0) = 0$ and $f'(z_0) > 0$.

That means that $f'(z_0)$ is real and positive. There is a theorem that holds that then there are two conformal mappings $h, \hat h:D \rightarrow G$ with $b \in D$ such that $h(b) = \hat h(b) = 0$. Is $h'(b)/\hat h'(b) > 0$ then $h = \hat h$.

So if you look at the conformal mapping $h \circ g$ where $g$ maps $D$ to a simply connected and bounded domain $\subset \mathbb{D}$ and $h$ expands $g(D)$ conformally to $\mathbb{D}$.

Using a rotation $\theta : \mathbb{D} \rightarrow \mathbb{D}$ around $0$ makes the conformal map $\theta \circ h \circ g$ unique.

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