What conformal mappings are there of the form $f: \mathbb{D} \mapsto D$, where $\mathbb{D} = B_1(0)$ is the unit disk and $D \in \mathbb{C}$ ?

If $D$ is a simply connected domain (ie open, connected, no holes) in the complex plane, but not the entire complex plane, then there is a conformal map (ie analytic, one-to-one, onto) of $D$ onto the open unit disk $\mathbb{D}$.

We say that "$D$ is conformally equivalent to $\mathbb{D}$".

To find a unique conformal mapping $\, f$ from $D$ to $\mathbb{D}$, we need to specify "3 real parameters".

For example,

• We can use Möbius transformation that maps the upper half plane $\mathbb{D}^+$ to $\mathbb{D}$, by examining the mappings from $0, 1, \infty$ to $1, i, -1$. This would lead to

  $$\begin{aligned} f(z) = \frac{-z + i}{z + i} \end{aligned}$$

  Don't forget to check.

• $f$ would map the first quadrant $Q$ of the complex plane to $\mathbb{D}^+$. Again, check the mappings from $0, i, \infty$ (ie. the imaginary axis) to the line through $1, 0, -1$ (ie. the real axis).
• $g(z) = z^2$ is injective and analytic in $Q$, and $g$ maps $Q$ conformally onto $D$. Should check.

Now how can we use the above to help construct the Riemann map $h$ from $\mathbb{D}^+$ to $D$ ? Try and you should find the solution:

$$\begin{aligned} h = f \circ g \circ f^{-1} \end{aligned}$$

More at Analysis of a Complex Kind.