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Notes on complex analysis

Ryan Tully-Doyle

Section 1.6 Linear fractional tranformations

A couple of you have asked me how to actually find the conformal maps that are guaranteed to exist by the Riemann Mapping Theorem. For a large class of domains, we can readily (for some value of readily) find these!

Definition 1.6.1.

A linear fractional transformation is a degree one complex rational function of the form
\begin{equation*} T(z) = \frac{az + b}{cz + d} \end{equation*}
for fixed constants \(a, b, c, d \in \C\text{,}\) and where we assume that \(ad - bc \neq 0\text{.}\)
What happens to \(T\) if \(ad - bc \neq 0\text{?}\)
We will use the shorthand LFT to stand for linear fractional transformation. From what we learned last term, it should be immediately obvious that an LFT has one pole of order 1 where \(cz + d = 0\text{.}\)
LFTs are the basic building blocks of conformal maps.
Let \(S = \frac{-dw + b}{cw - a}\text{.}\) The domain \(A\) excludes the simple pole of \(T\text{,}\) and likewise for \(B\) and \(S\text{,}\) which means that \(T, S\) are analytic on \(A, B\) respectively. So the argument comes down to showing that these functions are inverses, which can be done by direct computation. We will need the fact that \(cw - a \neq 0\) on \(B\) and that \(bc - ad \neq 0\) by definition. We leave to the reader the computation that
\begin{equation*} T(S(w)) = w \end{equation*}
and
\begin{equation*} S(T(z)) = z\text{.} \end{equation*}
Finally,
\begin{equation*} 1 = \frac{d}{dz}(z) = \frac{d}{dz} (S(T(z))) = S\dd(T(z)) \cdot T\dd(z), \end{equation*}
which implies that \(T\dd(z) \neq 0.\)
We leave a little hook for the future here - we can “complete” the function by assigning the value \(T(-d/c) = \infty\text{,}\) which will turn out, which a little caution, to give us a conformal bijection of the Riemann sphere to itself. We'll arrive at the Riemann sphere in the next section.
What sort of functions are in the class of LFTs? If \(a = 1, c = 0, d=0\text{,}\) we get the translation function \(T(z) = z + b\text{.}\) If \(b = c = 0\) and \(d = 1\text{,}\) then we get \(T(z) = az\text{,}\) which should be viewed as \(T(z) = re^{i\theta}z\) which is a rotation by \(\theta\) and a magnification by \(r\text{.}\) Finally, if \(b=c=1\) and \(a=d=0\text{,}\) we get the inversion function \(T(z) = \frac{1}{z}\text{.}\)
We noted that the conformal automorphisms of the unit disk form a group - let us characterize these functions.
We'll start by showing that maps of the form of \(T\) map the unit circle into itself. First, assume that \(\abs{z} = 1\text{.}\) Then
\begin{equation*} \abs{T(z)} = \abs{\frac{z - \alpha}{1 - \cc\alpha z}} = \frac{1}{\abs{z}} \abs{\frac{z - \alpha}{z\inv - \cc\alpha}}. \end{equation*}
Because \(\abs{z} = 1\text{,}\) we also have that \(z\inv = \cc z\text{,}\) which gives
\begin{equation*} \abs{T(z)} = \frac{\abs{z - \alpha}}{\abs{\cc z - \cc \alpha}} = 1. \end{equation*}
Hence, \(T(\T) \subset \T\text{.}\) Now, note that \(T\) has only a singularity at \(z = \frac{1}{\cc\alpha}\text{,}\) which is outside the unit circle, so \(T\) is analytic in a neighborhood of \(\D\text{.}\) Then by the maximium modulus principle, for any \(z\in \D\text{,}\) we must have \(\abs{T(z)} \lt 1\text{,}\) which implies that \(T:\D \to \D\text{.}\) Also, by Proposition 1.6.3, \(T\inv\) has the same form, and so also defines a conformal map from \(\D \to \D\text{.}\) Thus any such \(T\) is a conformal automorphism from \(\D \to \D\text{.}\)
Now, let's show that any conformal automorphism from \(\D \to \D\) must be of this form. We'll be using uniqueness of conformal maps. Assume that \(R:\D \to \D\) is a conformal automorphism. Let \(\alpha = R\inv(0)\text{,}\) and let \(\theta = \operatorname{arg} R\dd(z_0)\text{.}\) A map \(T\) of the form (1.6.1) also has \(T(\alpha) = 0\text{.}\) Further, if we compute the derivative of \(T\) at \(\alpha\text{,}\) we'll get
\begin{equation*} T\dd(\alpha) =e^{i\theta} \frac{1}{1 - \abs{\alpha}^2}\text{,} \end{equation*}
which has argument \(\operatorname{arg} T\dd(\alpha) = \theta\text{.}\) Thus, \(R\) and \(T\) agree in value and argument at \(\alpha\text{,}\) and so by uniquess of conformal maps, \(R = T\text{.}\) (note: the argument version of uniqueness follows from the positive derivative version. add this to a future draft or add as exercise.)
Let \(\hat\C\) denote the extended complex plane. If we think of lines as passing through the unique point at \(\infty\) in either direction, then lines are essentially circles in \(\hat\C\text{.}\) Equivalently, lines map to great circles passing through the north pole in the Riemann sphere model. Linear fractional transforms on \(\hat\C\) send circles to circles with this identification for lines.
Some authors use the amusingly awful word “clircle” to include both cases. We'll be more dignified and just say “circle” when we're referring to these shapes in the extended plane or sphere.
Now suppose that we are given three points in \(\C\text{.}\) If these are collinear, the three points determine a unique line. If they are not collinear, they determine a unique circle. Because LFTs take circles/lines to circles lines, we might suspect that the implication of the geometric fact above is that there is a unique linear fractional map carrying any set \(z_1, z_2, z_3 \in \C\) to the set \(w_1, w_2, w_3 \in \C\text{,}\) and this is indeed the case.
Figure 1.6.7. Useful conformal maps
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