# Riemann mapping theorem. Why using rotations?

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?

• I cannot see any reason for using $\theta$. In fact, $h \circ g$ is conformal. – Paul Frost Oct 23 '18 at 15:39
• 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: $f'(z_0) > 0$ – Arjihad Oct 23 '18 at 18:54

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.