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Complex Analysis: an Open Source Textbook

Russell W. Howell, John H. Mathews

Section 9.2 Bilinear Transformations

Another important class of elementary mappings was studied by Augustus Ferdinand Möbius (1790–1868). These mappings are conveniently expressed as the quotient of two linear expressions. They arise naturally in mapping problems involving the function \(\mathrm{Arctan}(z)\text{.}\) In this section, we show how they are used to map a disk one-to-one and onto a half-plane.
If we let \(a, \, b, \, c\text{,}\) and \(d\) denote four complex constants with the restriction that \(ad\ne bc\text{,}\) then the function
\begin{equation} w=S(z)=\frac{az+b}{cz+d}\tag{9.2.1} \end{equation}
is called a bilinear transformation, a Möbius transformation, or a linear fractional transformation. If the expression for \(S\) in Equation (9.2.1) is multiplied by the quantity \(cz+d\text{,}\) then the resulting expression has the bilinear form \(cwz-az+dw-b=0\text{.}\) We collect terms involving \(z\) and write \(z(cw-a)=-dw+b\text{.}\) Then, for values of \(w\ne \frac{a}{c}\text{,}\) the inverse transformation is given by
\begin{equation} z=S^{-1}(w) =\frac{-dw+b}{cw-a}\text{.}\tag{9.2.2} \end{equation}
We can extend \(S\) and \(S^{-1}\) to mappings in the extended complex plane. The value \(S(\infty)\) should equal the limit of \(S(z)\) as \(z \to \infty\text{.}\) Therefore we define
\begin{equation*} S(\infty) = \lim_{z \to \infty} S(z) = \lim_{z \to \infty} \frac{a+\frac{b}{z}}{c+\frac{d}{z}}=\frac{a}{c}\text{,} \end{equation*}
and the inverse is \(S^{-1}(\frac{a}{c})=\infty\text{.}\) Similarly, the value \(S^{-1}(\infty)\) is obtained by
\begin{equation*} S^{-1}(\infty) = \lim_{w \to \infty}\ S^{-1}(w) = \lim_{w \to \infty} \frac{-d+\frac{b}{w}}{c-\frac{a}{w}}=\frac{-d}{c}\text{,} \end{equation*}
and the inverse is \(S(\frac{-d}{c})=\infty\text{.}\) With these extensions we conclude that the transformation \(w=S(z)\) is a one-to-one mapping of the extended complex \(z\) plane onto the extended complex \(w\) plane.
We now show that a bilinear transformation carries the class of circles and lines onto itself. If \(S\) is an arbitrary bilinear transformation given by Equation (9.2.1) and \(c=0\text{,}\) then \(S\) reduces to a linear transformation, which carries lines onto lines and circles onto circles. If \(c\ne 0\text{,}\) then we can write \(S\) in the form
\begin{equation} S(z) = \frac{a(cz+d)+bc-ad}{c(cz+d)} = \frac{a}{c}+\left(\frac{bc-ad}{c}\right)\!\left(\frac{1}{cz+d}\right)\text{.}\tag{9.2.3} \end{equation}
The condition \(ad \ne bc\) precludes the possibility that \(S\) reduces to a constant. Equation (9.2.3) indicates that \(S\) can be considered as a composition of functions. It is a linear mapping \(\xi =cz+d\text{,}\) followed by the reciprocal transformation \(Z=\frac{1}{\xi }\text{,}\) followed by \(w=\frac{a}{c}+\frac{bc-ad}{c}Z\text{.}\) In Chapter 2 we showed that each function in this composition maps the class of circles and lines onto itself; it follows that the bilinear transformation \(S\) has this property. A half-plane can be considered to be a family of parallel lines and a disk as a family of circles. Therefore we conclude that a bilinear transformation maps the class of half-planes and disks onto itself. Example 9.2.1 illustrates this idea.

Example 9.2.1.

Show that \(w=S(z)=\frac{i(1-z)}{1+z}\) maps the unit disk \(\left\vert z\right\vert \lt 1\) one-to-one and onto the upper half-plane \(\mathrm{Im}(w) >0\text{.}\)
Solution.
We first consider the unit circle \(C:|z|=1\text{,}\) which forms the boundary of the disk and find its image in the \(w\) plane. If we write \(S(z)=\frac{-iz+i}{z+1}\text{,}\) then we see that \(a=-i, \, b=i \, c=1\text{,}\) and \(d=1\text{.}\) Using Equation (9.2.2), we find that the inverse is given by
\begin{equation} z=S^{-1}(w)=\frac{-dw+b}{cw-a}=\frac{-w+i}{w+i}\text{.}\tag{9.2.4} \end{equation}
If \(|z|=1\text{,}\) then Equation (9.2.4) implies that the images of points on the unit circle satisfy the equation
\begin{equation} |w+i| = |-w+i|\text{.}\tag{9.2.5} \end{equation}
Squaring both sides of Equation (9.2.5), we obtain \(u^2+(1+v)^2=u^2+(1-v)^2\text{,}\) which can be simplified to yield \(v=0\text{,}\) which is the equation of the \(u\)-axis in the \(w\) plane.
The circle \(C\) divides the \(z\) plane into two portions, and its image is the \(u\)-axis, which divides the \(w\) plane into two portions. The image of the point \(z=0\) is \(w=S(0) =i\text{,}\) so we expect that the interior of the circle \(C\) is mapped onto the portion of the \(w\) plane that lies above the \(u\)-axis. To show that this outcome is true, we let \(|z| \lt 1\) . Then Equation (9.2.4) implies that the image values must satisfy the inequality \(|-w+i| \lt |w+i|\text{,}\) which we write as
\begin{equation*} d_1 = |w-i| \lt |w-(-i)| = d_2\text{.} \end{equation*}
If we interpret \(d_1\) as the distance from \(w\) to \(i\) and \(d_2\) as the distance from \(w\) to \(-i\text{,}\) then a geometric argument shows that the image point \(w\) must lie in the upper half-plane \(\mathrm{Im}(w)>0\text{,}\) as shown in Figure 9.2.2. As \(S\) is one-to-one and onto in the extended complex plane, it follows that \(S\) maps the disk onto the half-plane.
Figure 9.2.2. The image of \(|z| \lt 1\) under \(w=\frac{i(1-z)}{1+z}\)
The general formula for a bilinear transformation (Equation (9.2.1)) appears to involve four independent coefficients: \(a, \, b, \, c\text{,}\) and \(d\text{.}\) But as \(S(z) \ne K\text{,}\) either \(a \ne 0\) or \(c \ne 0\text{,}\) we can express the transformation with three unknown coefficients and write either
\begin{equation*} S(z) = \frac{z+\frac{b}{a}}{\frac{cz}{a}+\frac{d}{a}}, \text{ or } S(z) = \frac{\frac{az}{c}+\frac{b}{c}}{z+\frac{d}{c}}\text{,} \end{equation*}
respectively. Doing so permits us to determine a unique a bilinear transformation if three distinct image values \(S(z_1)=w_1, \, S(z_2)=w_2\text{,}\) and \(S(z_3)=w_3\) are specified. To determine such a mapping, we can conveniently use an implicit formula involving \(z\) and \(w\text{.}\)

Proof.

We algebraically manipulate Equation (9.2.6) and solve for \(w\) in terms of \(z\text{.}\) The result is an expression for \(w\) that has the form of Equation (9.2.1), where the coefficients \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) involve various combinations of the values \(z_1, \, z_2, \, z_3, \, w_1, \, w_2\text{,}\) and \(w_3\text{.}\) The details are left as an exercise.
If we set \(z=z_1\) and \(w=w_1\) in Equation (9.2.6), then both sides of the equation are zero, showing that \(w_1\) is the image of \(z_1\text{.}\) If we set \(z=z_2\) and \(w=w_2\) in Equation (9.2.6), then both sides of the equation take on the value \(1\text{.}\) Hence \(w_2\) is the image of \(z_2\text{.}\) Taking reciprocals, we write Equation (9.2.6) in the form
\begin{equation} \left(\frac{z-z_3}{z-z_1}\right)\!\left(\frac{z_2-z_1}{z_2-z_3}\right) = \left(\frac{w-w_3}{w-w_1}\right)\!\left(\frac{w_2-w_1}{w_2-w_3}\right)\text{.}\tag{9.2.7} \end{equation}
If we set \(z=z_3\) and \(w=w_3\) in Equation (9.2.7), then both sides of the equation are zero. Therefore \(w_3\) is the image of \(z_3\text{,}\) and we have shown that the transformation has the required properties.

Example 9.2.4.

Construct the bilinear transformation \(w=S(z)\) that maps the points \(z_1=-i\text{,}\) \(z_2=1\text{,}\) and \(z_3=i\) onto the points \(w_1=-1\text{,}\) \(w_2=0\text{,}\) and \(w_3=1\text{,}\) respectively.
Solution.
We use the implicit formula (Equation (9.2.6)) and write
\begin{equation*} \left(\frac{z+i}{z-i}\right)\!\left(\frac{1-i}{1+i}\right) = \left(\frac{w+1}{w-1}\right)\!\left(\frac{0-1}{0+1}\right) = \frac{w+1}{-w+1} \end{equation*}
Expanding this equation, we obtain
\begin{align*} (1+i) zw+(1-i)w \amp +(1+i)z +(1-i)\\ \amp = (-1+i) zw+(-1-i)w + (1-i)z+(1+i) \text{.} \end{align*}
Then, collecting terms involving \(w\) and \(zw\) on the left results in
\begin{equation*} 2w+2zw=2i-2iz\text{,} \end{equation*}
from which we obtain \(w(1+z) =i(1-z)\text{.}\) Therefore the desired bilinear transformation is
\begin{equation*} w=S(z) =\frac{i(1-z)}{1+z} \end{equation*}

Example 9.2.5.

Find the bilinear transformation \(w=S(z)\) that maps the points \(z_1=-2, \, z_2 = -1-i\text{,}\) and \(z_3=0\) onto \(w_1=-1, \, w_2=0\text{,}\) and \(w_3=1\text{,}\) respectively.
Solution.
Again, we use the implicit formula and write
\begin{equation*} \left(\frac{z-(-2)}{z-0}\right)\!\left(\frac{-1-i-0}{-1-i-(-2)}\right) = \left(\frac{w-(-1)}{w-1}\right)\!\left(\frac{0-1}{0-(-1)}\right)\text{.} \end{equation*}
Using the fact that \(\frac{-1-i}{1-i}=\frac{1}{i}\text{,}\) we rewrite this equation as
\begin{equation*} \frac{z+2}{iz}=\frac{1+w}{1-w}\text{.} \end{equation*}
We now expand the equation and obtain \(z+2-zw-2w=iz+izw\text{,}\) which can be solved for \(w\) in terms of \(z\text{,}\) giving the desired solution
\begin{equation*} w=S(z) =\frac{(1-i) z+2}{(1+i) z+2} \end{equation*}
We let \(D\) be a region in the \(z\) plane that is bounded by either a circle or a straight line \(C\text{.}\) We further let \(z_1\text{,}\) \(z_2\text{,}\) and \(z_3\) be three distinct points that lie on \(C\) and have the property that an observer moving along \(C\) from \(z_1\) to \(z_3\) through \(z_2\) finds the region \(D\) to be on the left. If \(C\) is a circle and \(D\) is the interior of \(C\text{,}\) then we say that \(C\) is positively oriented. Conversely, the ordered triple \((z_1,z_2,z_3)\) uniquely determines a region that lies to the left of \(C\text{.}\)
We let \(G\) be a region in the \(w\) plane that is bounded by either a circle of a straight line \(K\text{.}\) We further let \(w_1\text{,}\) \(w_2\text{,}\) and \(w_3\) be three distinct points that lie on \(K\) such that an observer moving along \(K\) from \(w_1\) to \(w_3\) through \(w_2\) finds the region \(G\) to be on the left. Because a bilinear transformation is a conformal mapping that maps the class of circles and straight lines onto itself, we can use the implicit formula to construct a bilinear transformation \(w=S(z)\) that is a one-to-one mapping of \(D\) onto \(G\text{.}\)

Example 9.2.6.

Show that the mapping
\begin{equation*} w=S(z) =\frac{(1-i) z+2}{(1+i) z+2} \end{equation*}
maps the disk \(D:\left\vert z+1\right\vert \lt 1\) onto the upper half-plane \(\mathrm{Im}(w)>0\text{.}\)
Solution.
For convenience, we choose the ordered triple \(z_1=-2\text{,}\) \(z_2=-1-i\text{,}\) and \(z_3=0\text{,}\) which gives the circle \(C : |z+1|=1\) a positive orientation and the disk \(D\) a left orientation. From Example 9.5, the corresponding image points are
\begin{equation*} w_1=S(z_1) =-1, w_2 = S(z_2) = 0, \text{ and } w_3 = S(z_3) = 1\text{.} \end{equation*}
Because the ordered triple of points \(w_1\text{,}\) \(w_2\text{,}\) and \(w_3\) lie on the \(u\)-axis, it follows that the image of circle \(C\) is the \(u\)-axis. The points \(w_1\text{,}\) \(w_2\text{,}\) and \(w_3\) give the upper half-plane \(G\text{:}\) \(\mathrm{ Im}(w) >0\) a left orientation. Therefore \(w=S(z)\) maps the disk \(D\) onto the upper half-plane \(G\text{.}\) To check our work, we choose a point \(z_0\) that lies in \(D\) and find the half-plane in which its image, \(w_0\text{,}\) lies. The choice \(z_0=-1\) yields \(w_0=S(-1) =i\text{.}\) Hence the upper half-plane is the correct image. This situation is illustrated in Figure 9.2.7.
Figure 9.2.7. The bilinear mapping \(w = S(z) =(1-i)z+2](1+i)z+2]\)

Proof.

Case 1 If \(z_3=\infty\text{,}\) then we can write \(\frac{z_2-z_3}{z-z_3}=\frac{z_2-\infty}{z-\infty}=1\) and substitute this expression into Equation (9.2.6) to obtain
\begin{equation*} \frac{z-z_1}{z_2-z_1}=\left(\frac{w-w_1}{w-w_3}\right)\!\left(\frac{w_2-w_3}{w_2-w_1}\right) \end{equation*}
Case 2 If \(w_3 = \infty\text{,}\) then we can write \(\frac{w_2-w_3}{w-w_3} = \frac{ w_2-\infty}{w-\infty} = 1\) and substitute this expression into Equation (9.2.6) to obtain
\begin{equation} \left(\frac{z-z_1}{z-z_3}\right)\!\left(\frac{z_2-z_3}{z_2-z_1}\right) = \frac{w-w_1}{ w_2-w_1}\text{.}\tag{9.2.8} \end{equation}
Equation (9.2.8) is sometimes used to map the crescent-shaped region that lies between the tangent circles onto an infinite strip.

Example 9.2.9.

Find the bilinear transformation that maps the crescent-shaped region that lies inside the disk \(|z-2|\lt 2\) and outside the circle \(|z-1|=1\) onto a horizontal strip.
Solution.
For convenience we choose \(z_1=4\text{,}\) \(z_2=2+2i\text{,}\) and \(z_3=0\) and the image values \(w_1=0\text{,}\) \(w_2=1\text{,}\) and \(w_3=\infty\text{,}\) respectively. The ordered triple \(z_1\text{,}\) \(z_2\text{,}\) and \(z_3\) gives the circle \(|z-2| =2\) a positive orientation and the disk \(| z-2| \lt 2\) has a left orientation. The image points \(w_1\text{,}\) \(w_2\text{,}\) and \(w_3\) all lie on the extended \(u\)-axis, and they determine a left orientation for the upper half-plane \(\mathrm{Im}(w) >0\text{.}\) Therefore we can use the second implicit formula given in Equation (9.2.8) to write
\begin{equation*} \left(\frac{z-4}{z-0}\right)\!\left(\frac{2+2i-0}{2+2i-4}\right)=\frac{w-0}{1-0}\text{,} \end{equation*}
which determines a mapping of the disk \(|z-2| \lt 2\) onto the upper half-plane \(\mathrm{Im}(w) >0\text{.}\) We simplify the preceding equation to obtain the desired solution:
\begin{equation*} w = S(z) = \frac{-iz+4i}{z} \end{equation*}
A straightforward calculation shows that the points \(z_4=1-i, \, z_5=2\text{,}\) and \(z_6=1+i\) are mapped onto the points
\begin{equation*} w_4 = S(1-i) = -2+i, w_5 = S(2) =, \text{ and } w_6 = S(1+i) = 2+i\text{,} \end{equation*}
respectively. The points \(w_4, \, w_5\text{,}\) and \(w_6\) lie on the horizontal line \(\mathrm{Im}(w) =1\) in the upper half-plane. Therefore the crescent-shaped region is mapped onto the horizontal strip \(0 \lt \mathrm{Im}(w) \lt 1\text{,}\) as shown in Figure 9.2.10.
Figure 9.2.10. The mapping \(w=S(z) =\frac{-iz+4i}{z}\)

Subsection 9.2.1 Lines of Flux

In electronics, images of certain lines represent lines of electric flux, which comprise the trajectory of an electron placed in an electrical field. Consider the bilinear transformation
\begin{equation*} w = S(z) = \frac{z}{z-a} \text{ and } z = S^{-1}(w) = \frac{aw}{w-1} \end{equation*}
The half rays \(\{\mathrm{Arg}(w)=c\}\text{,}\) where \(c\) is a constant, that meet at the origin \(w=0\) represent the lines of electric flux produced by a source located at \(w=0\) (and a sink at \(w=\infty\)). The preimage of this family of lines is a family of circles that pass through the points \(z=0\) and \(z=a\text{.}\) We visualize these circles as the lines of electric flux from one point charge to another. The limiting case as \(a\to 0\) is called a dipole. The graphs for \(a=1, \, a=0.5\text{,}\) and \(a=0.1\) are shown in Figure 9.2.11.
Figure 9.2.11. Images of \(\mathrm{Arg}(w) = c\) under the mapping \(z = \frac{aw}{w-1}\)

Exercises Exercises

1.
If \(w=S(z) =\frac{(1-i) z+2}{(1+i) z+2}\text{,}\) find \(S^{-1}(w)\text{.}\)
Solution.
\(z=S^{-1}(w) = \frac{-2w+2}{(1+i)w-(1-i)} = \frac{(1-i)(1-w)}{i+w}\text{.}\)
2.
If \(w=S(z) =\frac{i+z}{i-z}\text{,}\) find \(S^{-1}(w)\text{.}\)
3.
Find the image of the right half-plane \(\mathrm{Re}(z) >0\) under \(w=\frac{i(1-z)}{1+z}\text{.}\)
Solution.
The disk \(|w|\lt 1\text{.}\)
4.
Show that the bilinear transformation \(w=\frac{i(1-z)}{1+z }\) maps the portion of the disk \(|z|\lt 1\) that lies in the upper half-plane \(\mathrm{Im}(z)>0\) onto the first quadrant \(u>0, \, v>0\text{.}\)
5.
Find the image of the upper half-plane \(\mathrm{Im}(z) >0\) under the transformation
\begin{equation*} w=\frac{(1-i) z+2}{(1+i) z+2} \end{equation*}
Solution.
The region \(|w|>1\text{.}\)
6.
Find the bilinear transformation \(w=S(z)\) that maps the points \(z_1=0, \, z_2=i\text{,}\) and \(z_3=-i\) onto \(w_1=-1, \,w_2=1\text{,}\) and \(w_3=0\text{,}\) respectively.
7.
Find the bilinear transformation \(w=S(z)\) that maps the points \(z_1=-i, \, z_2=0\text{,}\) and \(z_3=i\) onto \(w_1=-1, \, w_2=i\text{,}\) and \(w_3=1\text{,}\) respectively.
Solution.
\(w=S(z)=\frac{-iz+i}{1+z}\text{.}\)
8.
Find the bilinear transformation \(w=S(z)\) that maps the points \(z_1=0, \, z_2=1\text{,}\) and \(z_3=2\) onto \(w_1=0, \, w_2=1\text{,}\) and \(w_3=\infty\text{,}\) respectively.
9.
Find the bilinear transformation \(w=S(z)\) that maps the points \(z_1=1, \, z_2=i\text{,}\) and \(z_3=-1\) onto \(w_1=0, \, w_2=1\text{,}\) and \(w_3=\infty\text{,}\) respectively.
Solution.
\(w=S(z)=\frac{1-iz}{1+z}\text{.}\)
10.
Show that the transformation \(w=\frac{i+z}{i-z}\) maps the unit disk \(|z|\lt 1\) onto the right half-plane \(\mathrm{Re}(w)>0\text{.}\)
11.
Find the image of the lower half-plane \(\mathrm{Im}(z)\lt 0\) under \(w = \frac{i+z}{i-z}\text{.}\)
Solution.
The disk \(|w|\lt 1\text{.}\)
12.
If \(S_1(z) =\frac{z-2}{z+1}\) and \(S_2(z) = \frac{z}{z+3}\text{,}\) find \(S_1\big(S_2(z)\big)\) and \(S_2\big(S_1(z)\big)\text{.}\)
13.
Find the image of the quadrant \(x>0\text{,}\) \(y>0\) under \(w=\frac{z-1}{z+1}\text{.}\)
Solution.
The portion of the disk \(|w|\lt 1\) that lies in the upper half-plane \(\mathrm{Im}(w)>0\text{.}\)
14.
Show that Equation (9.2.6) can be written in the form of Equation (9.2.1).
15.
Find the image of the horizontal strip \(0\lt y\lt 2\) under \(w=\frac{z}{z-i}\text{.}\)
Solution.
The region of that lies exterior to both the circles \(|w-\frac{1}{2}| = \frac{1}{2}\) and \(|w-\frac{3}{2}| = \frac{1}{2}\text{.}\)
16.
Show that the bilinear transformation \(w=S(z) =\frac{az+b}{cz+d}\) is conformal at all points \(z \ne \frac{-d}{c}\text{.}\)
17.
A fixed point of a mapping \(w=f(z)\) is a point \(z_0\) such that \(f(z_0)=z_0\text{.}\) Show that a bilinear transformation can have at most two fixed points.
Solution.
The equation \(z=\frac{az+ib}{cz+d}\) can be written as \(cz^2+(d-a)z-b=0\text{,}\) and a quadratic equation has, at most, two distinct solutions.
18.
Find the fixed points of
(a)
\(w=\frac{z-1}{z+1}\text{.}\)
(b)
\(w=\frac{4z+3}{2z-1}\text{.}\)
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