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.