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# 18.8: Elementary transformations

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The following three types of fractional linear transformations are called elementary:

1. $$z \mapsto z + w,$$

2. $$z \mapsto w \cdot z$$ for $$w \ne 0,$$

3. $$z\mapsto \dfrac{1}{z}.$$

The geometric interpretation

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 $$\mathbb{R}$$.

Proposition $$\PageIndex{1}$$

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

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

$$f(z) = \frac{a\cdot z + b}{d}.$$

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}

Corollary $$\PageIndex{1}$$

The image of a circline under a fractional linear transformation is a circline.

Proof

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.

Exercise $$\PageIndex{1}$$

Show that the inverse of a fractional linear transformation is a fractional linear transformation.

Hint

Show that the inverse of each elementary transformation is elementary and use Proposition $$\PageIndex{1}$$.

Exercise $$\PageIndex{2}$$

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

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 $$g = f$$.

Exercise $$\PageIndex{3}$$

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

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,

$$z' = \overline{(\dfrac{\bar{w} \cdot z + [r^2 - |w|^2]}{z - w})}$$.