Complex Analysis II, Michaelmas 2024. Assignment 6: Möbius Transformations I

(i) We have \(M_{T}\circ M_{S}(z)=\displaystyle{\frac{(ap+br)z+(aq+bs)}{(cp+dr)z+(cq+ds)}}\), but also \[\left(\begin{array}{cc} ap+br&aq+bs\\ cp+dr&cq+ds \end{array}\right)=\left( \begin{array}{cc} a&b\\ c&d\end{array}\right).\left( \begin{array}{cc} p&q\\ r&s\end{array}\right) =T{S}.\]

(ii) \(M_{T} = {\rm Id}\) means \(z= M_T(z) = \frac{az+b}{cz+d}\) for every \(z \in \mathbb{C}\), so \(c z^2 + (d-a)z -b = 0\). Since this must hold for every \(z\) it must in particular hold for \(z=0\), and so we have \(b=0\). It must hold for \(z=1\) so \(c=a-d\) and also \(z=-1\) so \(c=-(a-d)\). This is only possible if \(c=a-d = 0\), whence \(T = \left(\begin{smallmatrix} a & 0 \\ 0 & a\end{smallmatrix}\right)\) with \(a \neq 0\) (since \(\det T \neq 0\)). The final claim follows immediately.

First note that \(-i\mapsto0\), \(i\mapsto\infty\) and \[1\mapsto\frac{1+i}{1-i}=\frac{2i}{2}=i.\] Thus the unit circle is sent to the circle or line through \(0,i,\infty\); i.e. the imaginary axis. And \(0\mapsto-1\), so the unit disc is sent to the left half-plane.

First note that \(i\mapsto\infty\) and \[\begin{aligned} 1&\mapsto\frac{2i}{1+i}=\frac{2i(1-i)}{2}=1+i, \quad &-1\mapsto\frac{-2}{1-i}=\frac{-2(1+i)}{2}=-1-i\ \end{aligned}\] so the unit circle is sent to the line or circle through \(\infty,1+i,-1-i\); in other words, the line \(u=v\) in the \((u,v)\)-plane (with angle \(\pi/4\)). And \(0\mapsto -1+i\), so the unit disc is sent to \(\left\{(u,v):v>u\right\}\). (Note that we could have used \(-i\mapsto0\) in place of one of the three points.)

First note that \(z=-1 \mapsto w=\infty\), so the unit circle is sent to a circle or line through \(\infty\); i.e. to a line.

Also, \(z=1\) maps to \(w=0\) whilst \(z=i\) maps to \[w=\dfrac{(1+i)i-1-i}{i+1}=\dfrac{-2}{i+1}=-(1-i)=-1+i,\] both of which lie on the line \(u+v=0\). Thus, the image of the unit circle must be this line. Finally, \(z=0\) is mapped to \[w=-1-i\] so the unit disc is sent to \(\left\lbrace (u,v):u+v < 0\right\rbrace\).

No. A Möbius transformation is a conformal map, so preserves angles between curves. The triangle with vertices at \(-1,i\) and \(1\) has angles \(\pi/4,\pi/4,\pi/2\) so cannot be mapped to an equilateral triangle whose angles are all equal to \(\pi/3\).

A simple calculation yields \[M_{C}\left(\frac{1}{2}(1+i)\right) \: = \: \frac{\frac{1}{2}(1+i) - i}{\frac{1}{2}(1+i) + i} \: = \: \frac{1-i}{1+3i} \: = \: \frac{(1-i)(1-3i)}{10} \: = \: \frac{-2-4i}{10}\: = \:-\frac 1 5(1+2i).\] The triangle in question is made up of three line segments. Each line segment must be taken to a line segment or circular arc by a Möbius transformation. Note that under \(M_{C}\) \[0 \mapsto \frac{-i}{i} = -1; \quad 1 \mapsto \frac{1-i}{1+i} = -i; \quad i \mapsto 0.\]

First we check where the line segment from \(0\) to \(1\) is taken. We have \[\frac 1 2 \mapsto \frac{\frac 1 2-i}{ \frac 1 2 + i} = \frac{1-2i}{1+2i}= \frac{(1-2i)^2}{5} = \frac{-3-4i}{5},\] which lies on the unit circle in the third quadrant. Thus the first line segment is taken to the circular arc in the third quadrant from \(-1\) to \(-i\).

Next we check the line segment from \(0\) to \(i\). We have \[\frac i 2 \mapsto \frac{\frac i 2-i}{ \frac i 2 + i} = \frac{-i}{3i}= -\frac{1}{3},\] which lies on the real axis. Thus, the second line segment is taken to the line segment joining \(-1\) and \(0\).

Finally, we consider the line segment between \(1\) and \(i\). The point \(\frac{1}{2}(1+i)\) lies on this line segment and is taken to \(-\frac 1 5(1+2i)\). We know that the image must be a line segment or a circular arc joining \(0\) and \(-i\), and since \(-\frac 1 5(1+2i)\) does not lie on the imaginary axis it must be a circular arc. Indeed, it must be the circular arc between \(0\) and \(-i\) in the third quadrant passing through \(0\) and \(-i\).

The image of the triangle is the union of the images of the three line segments considered; starting from the origin, this is a line from \(0\) to \(-1\), then a circular arc to \(-i\) (following the unit circle), then a circular arc in the third quadrant joining \(-i\) back to \(0\). So it looks like a fang!

If \(z_0\) is a fixed point it must satisfy \[\frac {iz_0+i}{-z_0+1} = z_0, \quad \text{ and so } \quad (z_0)^2-(1-i)z_0 + i = 0.\] Solving this (by completing the square or otherwise) yields \[z_0 = \left(\frac 1 2 \pm \frac{\sqrt 3}{2}\right ) - i\left(\frac 1 2 \pm \frac{\sqrt 3}{2} \right).\]

Let \(z_1\) be a point different from \(\alpha\) and \(\beta\), and let \(w_1\) be the image of \(z_1\) under the Möbius transformation. We know the Möbius transformation preserves the cross-ratio so (taking \(z_2=w_2=\alpha\) and \(z_3=w_3 = \beta\)) we have \[\frac{(f(z)-\alpha)(w_1-\beta)}{(f(z)-\beta)(w_1-\alpha)} =\frac{(z-\alpha)(z_1-\beta)}{(z-\beta)(z_1-\alpha)}.\] Hence \[\frac{f(z)-\alpha}{f(z)-\beta} =K\frac{z-\alpha}{z-\beta},\] where \(K= \frac{(w_1-\alpha)(z_1-\beta)}{(w_1-\beta)(z_1-\alpha)}\).

Consider \[\begin{equation} \label{eq:points} z_1=-i,z_2=-1,z_3=i,\qquad w_1=-i,w_2=0,w_3=i \end{equation}\]

We know that Möbius transformations preserve the cross-ratio; i.e., \[\frac{(w-w_2)(w_1-w_3)}{(w-w_3)(w_1-w_2)}=\frac{(z-z_2)(z_1-z_3)}{(z-z_3)(z_1-z_2)}.\] We simply need to solve for \(w\) in terms of \(z\). In this case, we have \[\begin{aligned} \frac{(w-0)(-i-i)}{(w-i)(-i-0)}=\frac{(z+1)(-i-i)}{(z-i)(-i+1)} \quad & \iff & \frac{-2iw}{-i(w-i)}&=\frac{-2i(z+1)}{(z-i)(-i+1)}\\ & \iff & w(z-i)(1-i)&=-i(w-i)(z+1)\\ & \iff &(1-i)wz-i(1-i)w&=-iwz-z-iw-1\\ & \iff &wz-w&=-z-1\\ & \iff &w(z-1)&=-z-1\\ & \iff &w&=-\dfrac{z+1}{z-1}. \end{aligned}\] Any three distinct points lie on a unique line or circle, so the Möbius transformation \(w = f(z)\) taking \(z_i\) into \(w_i\) for each \(i\) must take the unit circle to the imaginary axis (as the three image points are on the imaginary axis, see \(\eqref{eq:points}\)). Alternatively, we can take each point on the unit circle as \(z=e^{it}\) and get \[f(e^{it})=-\frac{e^{it}+1}{e^{it}-1}=-\frac{e^{it/2}(e^{it/2}+e^{-it/2})}{e^{it/2}(e^{it/2}-e^{-it/2})}=i\frac{\cos(t/2)}{\sin(t/2)}\] which goes along the imaginary axis if we vary \(t\in(-\pi,\pi]\), so this proves again that the unit circle is mapped to the imaginary axis. Choosing \(z=0\) in the interior of the unit disc, we see that its image \(f(z)=1\) under this map is in the right half-plane, so the interior of the circle must be sent to the right half plane \(\mathbb{H}_R:=\{ w \in \mathbb{C}: {\rm Re}(w) > 0 \}\). Note that \(z=1\) is sent to \(w=\infty\).

Here is a different way of solving this type of problem via linear algebra: We need to find complex numbers \(a\), \(b\), \(c\), \(d\) so that \[1= {a(-1)+b\over c(-1)+d}\ , \quad ie \quad -c+d=-a+b\, .\qquad \qquad \qquad (1)\] \[-1= {a(1)+b\over c(1)+d}\ , \quad ie \quad -c-d=a+b\, .\qquad \qquad \qquad (2)\] \[0= {a(-i)+b\over c(-i)+d}\ , \quad ie \quad -ai+b=0\, .\qquad \qquad \qquad (3)\]

Three linear equations in 4 unknowns, so 1 degree of freedom. Let’s choose \(a=1\). Then \((3)\) gives \(b=i\). Then \((1)+(2)\) gives \(-2c=2b\), ie, \(c=-i\), so \((2)\) then gives \(d=-c-a-b\), ie \(d=i-1-i=-1\).

So, required Möbius transformation is \[f(z)={z+i\over -iz-1}\ .\]

You can check this by substituting the given values [e.g., put \(z=-1\) and get \(f(-1)=(-1+i)/(i-1)= (-1+i)(-i-1)/2 =1\)].

We note that if \(z=-i\) then \(f(-i)=0\), so the unit circle is mapped to the real axis. Also, since \(z=0\) gives \(f(0)=-i\), we see that the unit disc is mapped to the lower half-plane.

We know that a Möbius transformation \(w=f(z)\) preserves the cross-ratio; i.e., \[\frac{(w-w_2)(w_1-w_3)}{(w-w_3)(w_1-w_2)}=\frac{(z-z_2)(z_1-z_3)}{(z-z_3)(z_1-z_2)}.\] We solve for \(w\) in terms of \(z\). In this case, the above simplifies to \[\begin{aligned} \frac{(w-(-i))(-1-1)}{(w-1)(-1-(-i))}& =\frac{(z-0)(-1+i-(1-i))}{(z-(1-i))(-1+i-0)}\\ -2(z-1+i)(w+i)&=(-2z+2iz)(w-1)\\ -(z-1+i)(w+i)&=z(i-1)(w-1)\\ -z+(1+i)&=(iz-1+i)w\\ w&=\dfrac{-z+1+i}{iz-1+i} \quad \left[ = \dfrac{iz+1-i}{z+1+i} \right]. %&=\dfrac{iz+1+i}{z-1+i}.\\ \end{aligned}\]

The boundary of \(\mathcal R\) is the line \(L=\left\{x+iy:x+y=0\right\}\), and the points \(-1+i,0,1-i\) all lie on this line. Thus, the Möbius transformation maps this line to the line or circle through \(-1,-i,1\). Clearly, this is the unit circle. We check what happens to \(\mathcal R\): \[z=1+i \quad \mapsto \quad w=\dfrac{i(1+i)+1-i}{1+i+1+i}=0,\] so \(\mathcal R\) is sent to the interior of the unit circle in the \(w\)-plane. Finally, \(z=\infty\) is mapped to \(w=\frac{i}{1}=i\).

We know that a Möbius transformation \(w=f(z)\) preserves the cross-ratio; i.e., \[\frac{(w-w_2)(w_1-w_3)}{(w-w_3)(w_1-w_2)}=\frac{(z-z_2)(z_1-z_3)}{(z-z_3)(z_1-z_2)}.\] We solve for \(w\) in terms of \(z\). In this case, the above simplifies to \[\frac{(w+i)(1-i)}{(w-i)(1+i)}=-\frac{z-(1+i)}{z+(1+i)}\ ,\] and the solution is \[w=\frac{z+i-1}{-z+i-1}\ .\]

The image of the line \(x=y\) is the unit circle, and, since 1 maps to \(i/(i-2)\) which clearly has modulus less than 1, the given region maps to the closed unit disc.

[Note that strictly speaking \(\mathcal{R}\) here is not a ‘region/domain’ since it is closed, so in this case the Möbius trans is not a biholomorphic map from \(\mathcal R\) to \(\overline B_1(0)\) by our definition.]

Assume \(|z|=1\). Then \[|z-z_0|=|z\bar z-z_0\bar z|=|1-z_0\bar z|=|\bar z_0z-1|,\] so \(|f(z)|=1\). Hence the Möbius transformation maps the unit circle to the unit circle. Clearly, \(z_0 \mapsto 0\), so the unit disc is mapped to the unit disc. We have \[(f \circ f)(z) \: = \: \frac{\frac{z-z_0}{\overline{z_0}\, z-1} - z_0}{\overline {z_0} \, \left(\frac{z-z_0}{\overline{z_0}\, z-1}\right) - 1} \: = \: \frac{z-z_0 - z_0(\overline{z_0}\, z-1)}{\overline {z_0} \, (z-z_0) - (\overline{z_0}\, z-1)}\: = \: \frac{z(1-z_0 \, \overline{z_0})}{(1-z_0 \, \overline{z_0})}\: = \: z,\] so \(f\) is its own inverse (it is an “involution”).

If \(|z_0|=1\), then the determinant of the matrix associated with \(f\) is \(``ad-bc"=-1-(-z_0\bar z_0)=0\), so we don’t have a Möbius transformation. If \(|z_0|>1\), then \(z_0 \mapsto 0\) so the Möbius transformation maps the exterior of the unit disc to the unit disc.