Möbius Transformations

Section 3.1 Möbius Transformations

The first class of functions that we will discuss in some detail are built from linear polynomials.

Definition 3.1.1.

A linear fractional transformation is a function of the form

\begin{equation*} f(z) \ = \ \frac{ az+b }{ cz+d } \end{equation*}

where \(a,b,c,d \in \C\text{.}\) If \(ad-bc \not= 0\) then \(f\) is called a Möbius
¹
Named after August Ferdinand Möbius (1790–1868).
transformation.

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Exercise 2.5.16 said that any polynomial in \(z\) is an entire function, and so the linear fractional transformation \(f(z) = \frac{ az+b }{ cz+d }\) is holomorphic in \(\C \setminus \{ - \frac d c \}\text{,}\) unless \(c=0\) (in which case \(f\) is entire). If \(c \ne 0\) then \(\frac{ az+b }{ cz+d } = \frac a c\) implies \(ad-bc = 0\text{,}\) which means that a Möbius transformation \(f(z) = \frac{ az+b }{ cz+d }\) will never take on the value \(\frac a c\text{.}\) Our first proposition in this chapter says that with these small observations about the domain and image of a Möbius transformation, we obtain a class of bijections, which are quite special among complex functions.

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Proposition 3.1.2.

Let \(a,b,c,d \in \C\) with \(c \ne 0\text{.}\) Then \(f: \C \setminus \{ - \frac d c \} \to \C \setminus \{ \frac a c \}\) given by \(f(z) = \frac{ az+b }{ cz+d }\) has the inverse function \(f^{-1}: \C \setminus \{ \frac a c \} \to \C \setminus \{ - \frac d c \}\) given by

\begin{equation*} f^{-1} (z) \ = \ \frac{ dz-b }{ -cz+a } \,\text{.} \end{equation*}

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If this reminds you of the inverse of a \(2 \times 2\) matrix, you should do Exercise 3.6.10.

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We remark that the formula for \(f^{-1} (z)\) in Proposition 3.1.2 works also when \(c = 0\text{,}\) except that in this case both domain and image of \(f\) are \(\C\text{;}\) see Exercise 3.6.2. In either case, we note that the inverse of a Möbius transformation is another Möbius transformation.

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Example 3.1.3.

Consider the linear fractional transformation \(f(z) = \frac{ z-1 }{ iz+i }\text{.}\) This is a Möbius transformation (check the condition!) with domain \(\C \setminus \{ -1 \}\) whose inverse can be computed via

\begin{equation*} \frac{ z-1 }{ iz+i } \ = \ w \qquad \Longleftrightarrow \qquad z \ = \ \frac{ iw + 1 }{ -iw + 1 } \, \text{,} \end{equation*}

so that \(f^{-1} (z) = \frac{ iz + 1 }{ -iz + 1 }\text{,}\) with domain \(\C \setminus \{ -i \}\text{.}\)

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Proof.

We first prove that \(f\) is one-to-one. If \(f(z_1) = f(z_2)\text{,}\) that is,

\begin{equation*} \frac{ az_1+b }{ cz_1+d } \ = \ \frac{ az_2+b }{ cz_2+d } \,\text{,} \end{equation*}

then \((az_1+b)(cz_2+d) = (az_2+b)(cz_1+d)\text{,}\) which can be rearranged to

\begin{equation*} (ad-bc)(z_1-z_2) \ = \ 0 \,\text{.} \end{equation*}

Since \(ad-bc \not= 0\) this implies that \(z_1 = z_2\text{.}\) This shows that \(f\) is one-to-one.

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Exercise 3.6.1 verifies that the Möbius transformation \(g(z) = \frac{ dz-b }{ -cz+a }\) is the inverse of \(f\text{,}\) and by what we have just proved, \(g\) is also one-to-one. But this implies that \(f: \C \setminus \{ - \frac d c \} \to \C \setminus \{ \frac a c \}\) is onto.

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We remark that Möbius transformations provide an immediate application of Proposition 2.2.6, as the derivative of \(f(z) = \frac{ az+b }{ cz+d }\) is

\begin{equation*} f'(z) \ = \ \frac{ a(cz+d) - c(az+b) }{ (cz+d)^2 } \ = \ \frac{ ad - bc }{ (cz+d)^2 } \end{equation*}

and thus never zero. Proposition 2.2.6 implies that Möbius transformations are conformal, that is, they preserve angles.

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Möbius transformations have even more fascinating geometric properties. En route to an example of such, we introduce some terminology. Special cases of Möbius transformations are translations \(f(z) = z + b\text{,}\) dilations \(f(z) = az\text{,}\) and inversion \(f(z) = \frac 1 z\text{.}\) The next result says that if we understand these three special Möbius transformations, we understand them all.

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Proposition 3.1.4.

Suppose \(f(z) = \frac{ az+b }{ cz+d }\) is a linear fractional transformation. If \(c=0\) then

\begin{equation*} f(z) \ = \ \frac a d \, z + \frac b d \,\text{,} \end{equation*}

and if \(c \not= 0\) then

\begin{equation*} f(z) \ = \ \frac{ bc-ad }{ c^2 } \frac 1 {z + \frac d c} + \frac a c \, \text{.} \end{equation*}

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In particular, every linear fractional transformation is a composition of translations, dilations, and inversions.

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Proof.

Simplify.

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Theorem 3.1.5.

Möbius transformations map circles and lines into circles and lines.

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Example 3.1.6.

Continuing Example 3.1.3, consider again \(f(z) = \frac{ z-1 }{ iz+i }\text{.}\) For \(\phi \in \R\text{,}\)

\begin{align*} f(e^{ i \phi }) \amp \ = \ \frac{ e^{ i \phi } - 1 }{ i \, e^{ i \phi } + i } \ = \ \frac{ \left( e^{ i \phi } - 1 \right) \left( e^{ -i \phi } + 1 \right) }{ i \left| e^{ i \phi } + 1 \right|^2 }\\ \amp \ = \ \frac{ e^{ i \phi } - e^{ -i \phi } }{ i \left| e^{ i \phi } + 1 \right|^2 } \ = \ \frac{ 2 \Im \left( e^{ i \phi } \right) }{ \left| e^{ i \phi } + 1 \right|^2 } \ = \ \frac{ 2 \sin \phi }{ \left| e^{ i \phi } + 1 \right|^2 } \,\text{,} \end{align*}

which is a real number. Thus Theorem 3.1.5 implies that \(f\) maps the unit circle to the real line.

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Proof.

Translations and dilations certainly map circles and lines into circles and lines, so by Proposition 3.1.4, we only have to prove the statement of the theorem for the inversion \(f(z) = \frac 1 z\text{.}\)

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The equation for a circle centered at \(x_0 + iy_0\) with radius \(r\) is \((x-x_0)^2 + (y-y_0)^2 = r^2\text{,}\) which we can transform to

\begin{equation} \alpha (x^2 + y^2) + \beta x + \gamma y + \delta \ = \ 0\tag{3.1} \end{equation}

for some real numbers \(\alpha\text{,}\) \(\beta\text{,}\) \(\gamma\text{,}\) and \(\delta\) that satisfy \(\beta^2 + \gamma^2 > 4 \, \alpha \delta\) (see Exercise 3.6.3). The form (3.1) is more convenient for us, because it includes the possibility that the equation describes a line (precisely when \(\alpha = 0\)).

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Suppose \(z = x+iy\) satisfies (3.1); we need to prove that \(u+iv := \frac 1 z\) satisfies a similar equation. Since

\begin{equation*} u + iv \ = \ \frac{ x - iy }{ x^2 + y^2 } \,\text{,} \end{equation*}

we can rewrite (3.1) as

\begin{align} 0 \amp \ = \ \alpha + \beta \frac{ x }{ x^2 + y^2 } + \gamma \frac{ y }{ x^2 + y^2 } + \frac \delta { x^2 + y^2 }\notag\\ \amp \ = \ \alpha + \beta u - \gamma v + \delta (u^2 + v^2) \,\text{.}\tag{3.2} \end{align}

But this equation, in conjunction with Exercise 3.6.3, says that \(u+iv\) lies on a circle or line.

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Figure 3.1.7 demonstrates the effect that the inversion \(f(z)=\frac1z\) has on horizontal and vertical lines. In particular, the vertical line defined by \(\Re(z)=x_0\) is mapped into the circle of radius \(\frac1{2x_0}\) centered at \(\left(\frac1{2x_0},0\right)\text{.}\)

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Figure 3.1.7. Inversion maps vertical lines, shown on the left, into the circles centered on the real axis. Horizontal lines are mapped into circles centered on the imaginary axis.
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