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\[w=f(z)=\frac{1}{z}.\]
It is a one-to-one mapping of the extended complex plane onto itself and turns the unit circle inside out. Observe that
\[w=f(z)=\frac{1}{r\,e^{i\phi}}=\frac{1}{r}\,e^{-i\phi}.\]Thus, if we set \(r\) fixed and \(\phi\) varies from \(0\) to \(2\,\pi\) then the circle of radius \(r\) centered at the origin is mapped onto a circle of radius \(\frac{1}{r}\) centered at the origin and traced backward. On the other hand, if we set \(\phi\) fixed and let the real number \(r\) vary from \(-\infty\) to \(\infty\) then the point at \(\infty\) goes to the origin and vice versa. A line passing through the origin making an angle \(\phi\) with the positive \(x\)-axis in the \(z\)-plane is mapped to a line passing through the origin making an angle \(-\phi\) with the positive \(x\)-axis in the \(w\)-plane.
However, it is wrong to conclude that inversion maps lines to lines and circles to circles. But it does the next best thing: the image of a line is either a line or a circle. By treating \(\infty\) as a point we could regard a line as a _generalized circle_ with radius \(\infty\) and can say an inversion maps generalized circles to generalized circles (one can prove this statement using stereographic projection). At any rate, it is not difficult to check whether the image will be a line or a circle.
* If the original line or circle passes through the origin, its image will contain the point at infinity, so it must be a line.
* If the original line or circle misses the origin, its image will be bounded, hence a circle.
**Definition 108**.: A linear fractional transformation (or Mobius transformation, or bilinear transformation) \(T:\tilde{\mathbb{C}}\to\tilde{\mathbb{C}}\) is a function of the form
\[T(z)=\frac{az+b}{cz+d}\]
with the restriction \(ad\neq bc\) and \(a,b,c,d\in\mathbb{C}\).
The following are some properties of linear fractional transformations that can be deduced from the definition.
* When \(c=0\), the Mobius transformation \[T(z)=\frac{az+b}{d}=\frac{a}{d}z+\frac{b}{d}\] is a linear transformation. If \(c\neq 0\) then after some algebra one can show \[T(z)=\frac{a}{c}+\frac{bc-ad}{c}\;\frac{1}{cz+d}.\] Hence \(T\) can be expressed as a composition of a finite sequence of translations, magnifications, rotation, and inversions.
* We may arrange these numbers \(a,b,c,d\in\mathbb{C}\) appearing in the above definition as a matrix \(A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\) for which \(\mathrm{det}A\neq 0\). Thus, \(A\) is invertible, in fact \(A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\ -c&a\end{bmatrix}\)Moreover, if \(\lambda\neq 0\), \(\lambda\in\mathbb{C}\), then the matrix \(B=\begin{bmatrix}\lambda a&\lambda b\\ \lambda c&\lambda d\end{bmatrix}\) gives rise to the same map.
* \(T^{\prime}(z)\neq 0\) for all \(z\neq-\dfrac{d}{c}\), \(T:\tilde{\mathbb{C}}\rightarrow\tilde{\mathbb{C}}\) is a one-to-one, onto map, and its inverse can be found either by setting \(w=T(z)=\dfrac{az+b}{cz+d}\) and solving for \(z\), or simply using the matrix \(\begin{bmatrix}d&-b\\ -c&a\end{bmatrix}\) to write \(z=T^{-1}(z)=\dfrac{dw-b}{-cw+a}\).
* From \(T(z)=\dfrac{az+b}{cz+d}\), we conclude \[T\left(\dfrac{-b}{a}\right)=0,\quad T\left(\dfrac{-d}{c}\right)=\infty,\quad \lim_{z\rightarrow\infty}T(z)=\dfrac{a}{c},\quad\text{and}\quad T(0)=\dfrac{b }{d}.\] The above set of equalities can be used to find a particular linear fractional transformation. For example, if we are asked to find the linear fractional transformation which maps the points \(0,1,\infty\) onto \(1,\infty,0\), we simply observe \(b=d,\quad c=-d\) because \(T(0)=\dfrac{b}{d}=1\) and \(T(1)=\infty\), so \(\dfrac{-d}{c}=1\). Moreover \(T(\infty)=\dfrac{a}{c}=0\) implies \(a=0\). Putting all of this information together will yield \(T(z)=\dfrac{1}{-z+1}\).
* Recall that a point \(z_{0}\) is a fixed point of \(T\) if \(T(z_{0})=z_{0}\). Given a linear fractional transformation \(T\), if \(\infty\) is a fixed point of \(T\), then there exists complex numbers \(a,b\) such that \(T(z)=az+b\). This follows from the fact that \(\lim_{z\rightarrow\infty}T(z)=\dfrac{a}{c}\), by hypothesis \(c=0\), and \[T(z)=\dfrac{az+b}{d}=\dfrac{a}{d}z+\dfrac{b}{d}.\] Perhaps it is not so straightforward to see that given a linear fractional transformation, if \(T\) has three fixed points then \(T\) is the identity (see Exercise 6 below).
### 3.6 Conformality of Mobius Transformation
If we think of the Mobius transformation as a map taking \(z\) to \(T(z)=\dfrac{az+b}{cz+d}\) with \(ad\neq bc\), then \(T^{\prime}(z)=\dfrac{ad-bc}{(cz+d)^{2}}\neq 0\) and therefore \(T\) is conformal in \(\mathbb{C}\setminus\left\{\frac{-d}{c}\right\}\) for \(c\neq 0\). However, we defined \(T:\tilde{\mathbb{C}}\to\tilde{\mathbb{C}}\) where \(\tilde{\mathbb{C}}=\mathbb{C}\cup\infty.\) Thus we need to extend the definition of conformality to such mappings. We say \(T\) is conformal at \(\infty\) if \(T^{*}(z)=T(1/z)\) is conformal at \(z=0\). If \(T\) maps \(\varsigma\in\mathbb{C}\) to \(\infty\), then we set \(S(z)=\frac{1}{T(z)}\) and we say \(T\) is conformal at \(\infty\) if \(S\) is conformal at \(\varsigma\). Another way to think about \(T\) being conformal at \(\infty\) is to define \(T^{*}\) as \(T^{*}(z)=f\left(\frac{1}{z}\right)\) and check if \(T^{*}\) is conformal at \(0\). With a little more work, one can show that the Mobius transformation has nonzero derivative at every point of \(\tilde{\mathbb{C}}\). Consider cases separately for \(c=0\) and \(c\neq 0\). For example, for the case \(c\neq 0\), since we know \(T(-d/c)=\infty\), by taking \(w=T(z)\) and letting \(\tau=1/w\), we get \(\tau=\frac{cz+d}{az+b}\) and this function has a nonzero derivative at \(z=0\).
**Theorem 130**.: _Given any three distinct points \(z_{1},\,z_{2},\,z_{3}\) on the Riemann sphere and three distinct points \(w_{1},\,w_{2},\,w_{3}\) there exists a unique fractional linear transformation \(T\) such that_
\[T(z_{i})=w_{i}\quad\text{for}\quad i=1,2,3.\]
Proof
: Suppose there are two linear fractional transformations \(T\) and \(S\) such that \(T(z_{i})=w_{i}\) and \(S(z_{i})=w_{i}\) for \(i=1,2,3\). Then the mapping
\[(T\circ S^{-1})(w_{i})=T(z_{i})=w_{i}\quad\text{for}\quad i=1,2,3,\]
i.e., \(T\circ S^{-1}\) has three fixed points, then \(T\circ S^{-1}=I\) or \(T=S\).
### Triplet Representation of Mobius Transformation
If we want to give an expression to a function which maps \(z_{1}\) to \(0\) and \(z_{2}\) to \(\infty\) then the function \(T(z)=\frac{z-z_{1}}{z-z_{2}}\) will do the job. Suppose we also want a third point \(z_{3}\) to be mapped to \(1\) as well, then the map
\[T(z)=\frac{z-z_{1}}{z-z_{2}}\,\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\]
is the right one. Actually, we will know more when we express \(T(z)\) in the next theorem.
**Theorem 131**.: _The function_
\[z\to\frac{z-z_{1}}{z-z_{2}}\,\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\]
_is the unique function such that \(T(z_{1})=0,T(z_{2})=\infty,\) and \(T(z_{3})=1\)._
**Definition 109**.: If \(w=T(z)\) is a function such that \(T(z_{i})=w_{i}\) for \(i=1,2,3\), then \(w\) and \(z\) are related by the _cross-ratio formula_ given as:
\[\frac{w-w_{1}}{w-w_{2}}\:\frac{w_{3}-w_{2}}{w_{3}-w_{1}}=\frac{z-z_{1}}{z-z_{2}} \:\frac{z_{3}-z_{2}}{z_{3}-z_{1}}.\]
**Remark 73**.: Notice that in case \(T(z_{1})=0\), \(T(z_{2})=1\) and \(T(z_{3})=\infty\), then the cross-ratio formula will be
\[\frac{z-z_{1}}{z-z_{2}}\:\frac{z_{3}-z_{2}}{z_{3}-z_{1}}.\]
Furthermore, the cross-ratio formula can be modified so that one can still use it when \(\infty\) is one of the prescribed points in the extended \(z\) or \(w\) plane. For example if we suppose \(z_{1}=\infty\), then
\[\lim_{z_{1}\to\infty}\frac{z-z_{1}}{z-z_{2}}\:\frac{z_{3}-z_{2}}{z_{3}-z_{1}} =\frac{z_{3}-z_{2}}{z-z_{2}}.\]
Thus the cross-ratio formula takes the form
\[\frac{w-w_{1}}{w-w_{2}}\:\frac{w_{3}-w_{2}}{w_{3}-w_{1}}=\frac{z_{3}-z_{2}}{z -z_{2}}.\]
**Example 104**.: Find a linear fractional transformation \(T\) which maps \(0,1,\infty\) onto \(-1,0,1\). In this case the triplet representation of \(T\) is:
\[\frac{w-w_{1}}{w-w_{2}}\:\frac{w_{3}-w_{2}}{w_{3}-w_{1}}=\frac{z-z_{1}}{z-z_{2}}\]
and therefore
\[\frac{w+1}{w}\cdot\frac{1-0}{2}=\frac{z-0}{z-1}\]
and \(w=\frac{z-1}{z+1}\).
### Some Standard Examples
In the following we give some examples to illustrate various isomorphisms; details are left to the reader.
1. The quarter disc and the half disc (Figure 3.25).
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* Upper half disc with a half-strip (Figure 3.29).
* The quarter disc and upper half-plane (Figure 3.30).
Note that the above example \(z\mapsto\left(\frac{1+z^{2}}{1-z^{2}}\right)^{2}\) is a composition of three maps. First consider the map \(z\mapsto z^{2}\) between the quarter disc and half disc in example (a), then compose by the map \(f(z)=\frac{1+z}{1-z}\) in example (c) followed by the map \(f(z)=z^{2}\) in example (b).
### Exercises
1. Find a conformal map taking \(G=\{z:\ 0<\arg z<\frac{\pi}{2},\ \ \ 0<|z|<1\}\) to \(D=\{z:\ |z|<1\}\).
2. Find a conformal map from \(\{z\in\mathbb{C}:\ |z|<1\), \(\text{Re}z>0\}\) onto \(D\{z\in\mathbb{C}:\ |z|<1\}\). 3. Let \(f(z)=\dfrac{z-1}{z+1}\). What is the image under \(f\) of 1. The real line? 4. The circle with center \(0\) radius \(2\)? 5. The imaginary axis?
4. Find a linear transformation mapping the circle \(|z|=1\) onto the circle \(|w-5|=3\) and taking the point \(z=i\) to \(w=2\).
5. Suppose the linear transformation is given by \(w=(1+i)z+(1-i)\). What is the image of the rectangular region defined by \(O=(0,0),A=1\) and \(B=2i\)?
6. A _fixed point_ of a function \(f(z)\) is a point \(z_{0}\) satisfying \(f(z_{0})=z_{0}\). Show that a Mobius transformation \(f(z)\) can have at most two fixed points in the complex plane unless \(f(z)\equiv z\).
7. Let \(T\) be a linear fractional transformation. Prove that if \(T\) has three fixed points then \(T\) is an identity map.
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