The purpose of this post is to acquire a deeper appreciation of Mobius transformations. Typically, one simply encounters these as maps \(\mathcal M:\textbf C\cup\{\infty\}\to\textbf C\cup\{\infty\}\) on the Riemann sphere \(\textbf C\cup\{\infty\}\) of the form \(\mathcal M(z):=\frac{az+b}{cz+d}\) for \(ad-bc\neq 0\) without enough emphasis being placed on just how incredibly general and versatile this form is. The following problems (followed by their solutions) are meant to expose the full scope of possibilities that come with Mobius transformations.
Problem #\(1\): Solve the equation \(\frac{5}{8}=\frac{11-3x}{4+7x}\) for \(x\in\textbf R\).
Solution #\(1\): Exploit the invariance under Mobius transformations of a suitable cross-ratio:
\[\frac{(x-0)(\infty-(-4/7))}{(x-\infty)(0-(-4/7))}=\frac{(5/8-11/4)(-3/7-\infty)}{(5/8-(-3/7))(11/4-\infty)}\]
\[-\frac{7x}{4}=\frac{-17/8}{59/56}\]
\[x=\frac{68}{59}\]
Problem #\(2\): Solve the equation \(\frac{5}{8}=\frac{11-3x}{3+7x}\) for \(x\in\textbf R\).
Solution #\(2\): The trick is that any traceless Mobius transformation with \(a=-d\) is an involution. So the answer is just:
\[x=\frac{11-3\times(5/8)}{3+7\times(5/8)}=\frac{73}{59}\]
Problem #\(3\): Redo Problem #\(1\) but using the trick in Problem #\(2\).
Solution #\(3\): Since \(\text{lcm}(3,4)=12\), one has:
\[\frac{5}{8}=\frac{11-3x}{4+7x}\Rightarrow\frac{4}{3}\times\frac{5}{8}=\frac{4}{3}\times\frac{11-3x}{4+7x}\]
So:
\[x=\frac{4}{3}\times\frac{11-3\left(\frac{4}{3}\times\frac{5}{8}\right)}{4+7\left(\frac{4}{3}\times\frac{5}{8}\right)}=\frac{68}{59}\]
In general, the Mobius group is isomorphic to \(\cong\text{PGL}_2(\textbf C)\). This is incredibly powerful because it provides an explicit bridge between complex analysis and linear algebra. In this language, the question of which Mobius transformations are involutions \(\mathcal M^2=1\) boils down to which \(2\times 2\) complex invertible matrices \(M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\) satisfy \(M^2=1\). By the Cayley-Hamilton theorem, any matrix with eigenvalues \(\lambda=\pm 1\) will satisfy such an equation…
What’s the connection with unitary quantum logic gates on qubits? Give examples with 1D scattering (which is similar to Fresnel equations). Even for certain classes of intuitive \(2\)D matrices like rotations, what do the corresponding Mobius maps say? And also is there any connection between evaluating a Mobius map on some \(z\) and multiplying the corresponding matrix by a C^2 vector?
Problem #\(4\): Decompose the following rational function into partial fractions:
\[\frac{x^2+15}{(x+3)^2(x^2+3)}\]
Solution #\(4\): Remember that partial fractions may be viewed as just Mobius transformations with \(a=0\) (or powers thereof). In this case, that means:
\[\frac{x^2+15}{(x+3)^2(x^2+3)}=\frac{R_1}{x+3}+\frac{R_2}{(x+3)^2}+\frac{R_3}{x-\sqrt{3}i}+\frac{R_4}{x+\sqrt{3}i}\]
where the residues \(R_1,R_2,R_3,R_4\) may be calculated efficiently using the residue theorem (one may optionally prefer one’s residues to be real, in which case the last \(2\) terms should be combined as \(\frac{\tilde R_3x+\tilde R_4}{x^2+3}\); however as will be clear below it is actually very easy to get \(\tilde R_3,\tilde R_4\in\textbf R\) by simply summing \(\frac{R_3}{x-\sqrt{3}i}+\frac{R_4}{x+\sqrt{3}i}\) once the complex residues \(R_3,R_4\in\textbf C\) have been obtained; partial fraction composition is much easier decomposition!). To begin, it is helpful to complexify \(x\mapsto z\) everywhere:
\[\frac{z^2+15}{(z+3)^2(z^2+3)}=\frac{R_1}{z+3}+\frac{R_2}{(z+3)^2}+\frac{R_3}{z-\sqrt{3}i}+\frac{R_4}{z+\sqrt{3}i}\]
Now, suppose one wished to extract the residue at one of the simple poles, say \(R_1\). Then pick any contour in \(\textbf C\) that encloses only the corresponding simple pole, in this case at \(z=-3\) (strictly speaking \(z=-3\) is a double pole of the relevant function but that’s a technicality as far as computing \(R_1\) is concerned), and compute the corresponding contour integral on both sides, normalized by \(2\pi i\):
\[\frac{1}{2\pi i}\oint dz\frac{z^2+15}{(z+3)^2(z^2+3)}=\frac{1}{2\pi i}\oint dz\left(\frac{R_1}{z+3}+\frac{R_2}{(z+3)^2}+\frac{R_3}{z-\sqrt{3}i}+\frac{R_4}{z+\sqrt{3}i}\right)\]
On the right-hand side of course only the Mobius transformation with the simple pole at \(z=-3\) survives, leaving behind its residue:
\[R_1=\frac{1}{2\pi i}\oint dz\frac{(z^2+15)/(z^2+3)}{(z+3)^2}\]
By the generalized Cauchy integral formula, this is:
\[R_1=\frac{1}{1!}\left(\frac{d}{dz}\right)_{z=-3}\frac{z^2+15}{z^2+3}=\frac{1}{2}\]
By changing the contour to enclose the simple poles at \(z=\pm\sqrt{3}i\), one can similarly compute the residues \(R_3=1/(\sqrt{3}i-3)\) and \(R_4=-1/(3+\sqrt{3}i)\) (here just the vanilla Cauchy integral formula will do). This immediately leads to the residues \(\tilde R_3=-1/2\) and \(\tilde R_4=1/2\).
Finally to compute the “residue” \(R_2\) at the \(z=-3\) double pole, one simply has to convert it into a simple pole by multiplying through the whole equation by a factor of \((z+3)\):
\[\frac{z^2+15}{(z+3)(z^2+3)}=R_1+\frac{R_2}{z+3}+\frac{R_3(z+3)}{z-\sqrt{3}i}+\frac{R_4(z+3)}{z+\sqrt{3}i}\]
And now contour-integrate this around \(z=-3\):
\[R_2=\frac{(-3)^2+15}{(-3)^2+3}=2\]