(i) To any matrix \(T=\left(
\begin{smallmatrix} a&b\\ c&d\end{smallmatrix}\right) \in {\rm
GL}_2(\mathbb{C})\) we associate the Möbius transformation \(M_T(z)=\frac{az+b}{cz+d}\).
Given two Möbius transformations \(M_{T}(z)=\frac{az+b}{cz+d}\) and \(M_{S}(z)=\frac{pz+q}{rz+s}\) determine
\(M_{T}\circ M_{S}\) by a direct
calculation and conclude \[M_{T} \circ M_{S}
= M_{TS}.\] (ii) Show that \[M_{T} =
{\rm Id} \quad \iff \quad T = t\left(\begin{smallmatrix} 1 & 0 \\ 0
& 1\end{smallmatrix}\right) \text{ \quad (for some $t \in
\mathbb{C}^\ast$)}.\] Conclude that under composition a Möbius
transformation \(M_T\) has inverse
\((M_T)^{-1} = M_{T^{-1}}\).
Find the image of the unit circle and the unit disc under the Möbius transformation \(f(z)=w =\frac{z+i}{z-i}.\)
Find the image of the unit circle and the unit disc under the Möbius transformation \(f(z)=w=\frac{(1+i)z+i-1}{iz+1}\).
Find the image of the unit circle and the unit disc under the transformation \(f(z) = w=\frac{(1+i)z-1-i}{z+1}\).
Is there a Möbius transformation which maps the sides of the triangle with vertices at \(-1,i\) and \(1\) to the sides of an equilateral triangle (all sides of equal length)? Either give an example of such a Möbius transformation, or explain why it is not possible.
Show that the Cayley Map \(M_{C}=w=\frac{z-i}{z+i}\) takes the point \(\frac{1}{2}(1+i)\) to the point \(-\frac 1 5(1+2i)\). Hence, or otherwise, sketch the image of the triangle with vertices at \(0,1\) and \(i\) under \(M_{C}\).
Find the fixed points of the inverse Cayley Map \(M_{C^{-1}}\) [that is, the Möbius transformation associated with the matrix \(C^{-1}=\left(\begin{smallmatrix} i & i \\ -1 & 1\end{smallmatrix}\right)\)].
If \(\alpha\) and \(\beta\) are the two fixed points of a Möbius transformation \(f(z)\), show that for all \(z\neq \alpha,\beta\) and \(f(z)\neq\infty\), we have \[\frac{f(z)-\alpha}{f(z)-\beta} =K\frac{z-\alpha}{z-\beta},\] where \(K\) is a constant.
Find the Möbius transformation taking the ordered set of points \(\{-i,-1,i \}\) to the ordered set of points \(\{-i,0,i\}\). What is the image of the unit disc under this map? Which point is sent to \(\infty\)?
Find the Möbius transformation taking the ordered set of points \(\{-1,1,-i\}\) to the ordered set of points \(\{1,-1,0\}\). What is the image of the unit disc under this Möbius transformation?
Find the Möbius transformation taking the ordered set of points \(\{-1+i, \, 0, \, 1-i\}\) to the ordered set of points \(\{-1, \, -i,\, 1\}\). What is the image of the region \(\mathcal R=\left\{x+iy:x+y>0\right\}\) under this Möbius transformation? What happens to \(\infty\) under this Möbius transformation?
Find the Möbius transformation taking the ordered set of points \(\{0,1+i, -1-i\}\) to the ordered set of points \(\{1,-i,i\}\). What is the image of the region \(\mathcal R=\left\{x+iy:x-y\ge 0\right\}\) under this Möbius transformation?
Let \(z_0\) be an arbitrary complex number with \(|z_0| < 1\). Show that the Möbius transformation \[f(z)=\frac{z-z_0}{\overline{z_0}\, z-1}\] maps the unit disc to the unit disc, and maps \(z_0\) to \(0\), and \(0\) to \(z_0\). Compute \(f \circ f\). What do you observe? What happens if \(|z_0|=1\)? What happens if \(|z_0|>1\)?