18.8: Elementary transformations
- Page ID
- 58352
The following three types of fractional linear transformations are called elementary:
- \(z \mapsto z + w,\)
- \(z \mapsto w \cdot z\) for \(w \ne 0,\)
- \(z\mapsto \dfrac{1}{z}.\)
Suppose that \(O\) denotes the point with the complex coordinate \(0\).
The first map \(z \mapsto z+w,\) corresponds to the so-called parallel translation of the Euclidean plane, its geometric meaning should be evident.
The second map is called the rotational homothety with the center at \(O\). That is, the point \(O\) maps to itself and any other point \(Z\) maps to a point \(Z'\) such that \(OZ'=|w| \cdot OZ\) and \(\measuredangle ZOZ'=\arg w\).
The third map can be described as a composition of the inversion in the unit circle centered at \(O\) and the reflection across \(\mathbb{R}\) (the composition can be taken in any order). Indeed, \(\arg z \equiv -\arg \dfrac{1}{z}\). Therefore,
\(\arg z=\arg (1/\bar z);\)
that is, if the points \(Z\) and \(Z'\) have complex coordinates \(z\) and \(1/\bar{z}\), then \(Z'\in[OZ)\). Clearly, \(OZ=|z|\) and \(OZ'=|1/\bar{z}|=\dfrac{1}{|z|}\). Therefore, \(Z'\) is the inverse of \(Z\) in the unit circle centered at \(O\).
Finally, \(\dfrac{1}{z}=\overline{(1/\bar{z})}\) is the complex coordinate of the reflection of \(Z'\) across \(\mathbb{R}\).
The map \(f:\hat{\mathbb{C}} \to \hat{\mathbb{C}}\) is a fractional linear transformation if and only if it can be expressed as a composition of elementary transformations.
- Proof
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the "only if" part. Fix a fractional linear transformation
\(f(z) = \dfrac{a\cdot z + b}{c\cdot z + d}\).
Assume \(c \ne 0\). Then
\(\begin{array} {rcl} {f(z)} & = & {\dfrac{a}{c} - \dfrac{a \cdot d - b \cdot c}{c \cdot (c \cdot z + d)} =} \\ {} & = & {\dfrac{a}{c} - \dfrac{a \cdot d - b \cdot c}{c^2} \cdot \dfrac{1}{z + \tfrac{d}{c}}} \end{array}\)
That is,
\[f(z)=f_4\circ f_3\circ f_2\circ f_1 (z),\]
where \(f_1\), \(f_2\), \(f_3\), and \(f_4\) are the following elementary transformations:
\(\begin{aligned} f_1(z)&= z+\tfrac dc, & f_2(z)&= \tfrac1z, \\ f_3(z)&= - \tfrac{a\cdot d-b\cdot c}{c^2} \cdot z, & f_4(z)&= z+\tfrac ac.\end{aligned}\)
If \(c=0\), then
In this case \(f(z)=f_2\circ f_1 (z)\), where
\(\begin{aligned} f_1(z)&= \tfrac ad\cdot z, & f_2(z)= z+\tfrac bd.\end{aligned}\)
"If" part. We need to show that by composing elementary transformations, we can only get fractional linear transformations. Note that it is sufficient to check that the composition of a fractional linear transformation
\(f(z) = \frac{a\cdot z + b}{c\cdot z + d}.\)
with any elementary transformation \(z\mapsto z+w\), \(z\mapsto w\cdot z\), and \(z\mapsto \tfrac1z\) is a fractional linear transformation.
The latter is done by means of direct calculations.
\(\begin{aligned} \frac{a\cdot (z+w) + b}{c\cdot (z+w) + d} &= \frac{a\cdot z + (b+a\cdot w)}{c\cdot z + (d+c\cdot w)}, \\ \frac{a\cdot (w\cdot z) + b}{c\cdot (w\cdot z) + d} &= \frac{(a\cdot w)\cdot z + b}{(c\cdot w)\cdot z + d}, \\ \frac{a\cdot \frac1z + b}{c\cdot \frac1z + d} &= \frac{b\cdot z + a}{d\cdot z + c}.\end{aligned}\)
The image of a circline under a fractional linear transformation is a circline.
- Proof
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By Proposition \(\PageIndex{1}\), it is sufficient to check that each elementary transformation sends a circline to a circline.
For the first and second elementary transformation, the latter is evident.
As it was noted above, the map \(z\mapsto\dfrac{1}{z}\) is a composition of inversion and reflection. By Theorem 10.3.2, the inversion sends a circline to a circline. Hence the result.
Show that the inverse of a fractional linear transformation is a fractional linear transformation.
- Hint
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Show that the inverse of each elementary transformation is elementary and use Proposition \(\PageIndex{1}\).
Given distinct values \(z_0,z_1,z_\infty\in \hat{\mathbb{C}}\), construct a fractional linear transformation \(f\) such that
\(f(z_0)=0, \quad f(z_1)=1 \quad\text{and}\quad f(z_\infty)=\infty.\)
Show that such a transformation is unique.
- Hint
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The fractional linear transformation
\(f(z) = \dfrac{(z_1 - z_{\infty}) \cdot (z - z_0)}{(z_1 - z_0) \cdot (z - z_{\infty})}\)
meets the condition.
To show the uniqueness, assume there is another fractional linear transformation \(g(z)\) that meets the condition. Then the composition \(h = g\circ f^{-1}\) is a fractional linear transformation; set \(h(z) = \dfrac{a \cdot z + b}{c \cdot z + d}\).
Note that \(h(\infty) = \infty\); therefore, \(c = 0\). Further, \(h(0) = 0\) implies \(b = 0\). Finally, since \(h(1) = 1\), we get that \(\dfrac{a}{d} = 1\). Therefore, \(h\) is the identity; that is, \(h(z) = z\) for any \(z\). It follows that \(g = f\).
Show that any inversion is a composition of the complex conjugation and a fractional linear transformation.
Use Theorem 14.5.1 to conclude that any inversive transformation is either fractional linear transformation or a complex conjugate to a fractional linear transformation.
- Hint
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Let \(Z'\) be the inverse of the point \(Z\). Assume that the circle of the inversion has center \(W\) and radius \(r\). Let \(z, z'\), and \(w\) denote the complex coordinate of the points \(Z\), \(Z'\), and \(W\) respectively.
By the definition of inversion, \(\arg (z - w) = \arg (z'- w)\) and \(|z - w| \cdot |z' - w| = r^2\). It follows that \(\bar{z}' - \bar{w}) \cdot (z - w) = r^2\). Equivalently,
\(z' = \overline{(\dfrac{\bar{w} \cdot z + [r^2 - |w|^2]}{z - w})}\).