When is a mobius transformation its own inverse?

Question

I was puzzeling trying to find the inverse of the mobius transformation

$$ f(z) \ = \ \frac{z + i}{iz+1} $$ and if I am correct (I can be wrong here) it is its own inverse $( f(f(z)) = z )$

Are there general rules to check if any mobius transformation is its own inverse ?

something like $$ f(z) \ = \ \frac{az+b}{cz+d} $$ Is its own inverse iff:

Answer 1

I think the most easy way is to just calculate it: \begin{align} f(f(z)) &= \frac{a f(z) + b}{c f(z) + d}\\ &= \frac{a(az+b)+b(cz+d)}{c(az+b)+d(cz+d)}\\ &= \frac{(a^2+bc)z+(ab+bd)}{(ac+cd)z+(bc+d^2)} \stackrel!= z \end{align} Clearly for this to hold true, you need \begin{align} ab+bd &= 0\\ ac+cd &= 0\\ a^2+bc &= bc+d^2 \end{align} Now the last equation can be simplified to $a=\pm d$. Thus we have three cases to consider:

• $a=d=0$

  In this case, the first two equations are automatically fulfilled. Since a common factor in the coefficients doesn't change the function, we can choose $c=1$, and get $$f(z) = \frac{b}{z}$$ It is easily verified that this indeed is its own inverse.

• $a=d\ne 0$.

  In this case, the first two equations reduce to $b=c=0$ and we just get the identity.

• $a=-d\ne 0$.

  Again, the first two equations are automatically fulfilled; now we can use the invariance under a common factor to set $a=1$, obtaining $f(z) = \frac{z+b}{cz-1}$ Note that the function you found is of this type, with $b=c=\mathrm i$.

  Also note that functions of the form $f(z)=\alpha-z$ are obtained this way by setting $b=-\alpha$ and $c=0$.

Answer 2

The Möbius transformation are isomorphic to the projective group $PSL(2,\mathbb{C})=SL(2,\mathbb{C})/\{\pm I\}$. The question is thus when a matrix $$M= \begin{pmatrix} a & b\\ c & d \end{pmatrix}$$ (with $a d-bc=1$) is its own inverse; i.e., $M^2 = \pm I$. A brute force calculation shows that this is equivalent to the system of equations $$\begin{align} a^2+ bc &= \pm 1, & a b + bd &=0,\\ ac+c d &= 0, & b c + d^2 &=\pm 1. \end{align}$$

The solutions are given by (we distinguish two cases)

1) $b=0$:

• if we choose the $+$ sign in the system of equations, we obtain $a=1$ ($a=-1$ is equivalent due to the fact that we have a projective group). The remaining equations lead to $d= 1$ and $c=0$ (case i). Alternatively, we have $d=-1$ and $c$ arbitrary (but this contradicts $ad-bc=1$).

• if we choose the $-$ sign in the system of equations, we obtain $a=i$ ($a=-i$ is in the same equivalence class). The remaining equations lead to $d=-i$ and $c$ arbitrary (case ii). Alternatively, we have $d=i$ and $c=0$ (but this contradicts $ad-bc=1$).

2) $b\neq 0$:

• if we choose the $+$ sign in the system of equations, we obtain $c= (1-a^2)/b$ and $d=-a$; because $ad-bc=-1$ we have a contradiction.

• On the other hand for $-$, we obtain the solution $c= -(a^2+1)/b$ and $d=-a$ (case iii).

So in conclusions, we only have the following mutually exclusive cases:

i) $a=1, b=0,c=0,d=1$; i.e., $$f(z) = z$$

ii) $a=i, b=0, d=-i$; i.e., $$f(z) =\frac{z}{c z-1}, \qquad c \in \mathbb{C}$$

iii) $c= -(a^2+1)/b$, $d=-a$, $b\neq0$ (with $0\leq\mathop{\rm arg}b<\pi$ to make the solution unique); i.e., $$ f(z) = \frac{a z +b}{-(a^2+1) z/b -a} = -\frac{b (a z+b)}{(a^2+1) z +a b}, \qquad a,b \in \mathbb{C}; b\neq 0;0\leq\mathop{\rm arg}b<\pi$$

Answer 3

A Möbius transformation $f$ (which is not the identity) has exactly one or two fixed points in the extended complex plane.

If $f$ has one fixed point $z_0$ then for some $a \in \Bbb C$, $$ f(z) = z + a $$ if $z_0 = \infty$ or $$ \frac{1}{f(z) - z_0} = \frac{1}{z - z_0} + a $$ otherwise. It is easy to see that $f \circ f$ cannot be the identity in this case.

If $f$ has two fixed points $z_1, z_2$ then for some $\lambda \in \Bbb C$, $$ f(z) - z_1 = \lambda (z - z_1) $$ if $z_2 = \infty$, and $$ \frac{f(z)-z_1}{f(z)-z_2} = \lambda \frac{z-z_1}{z-z_2} $$ otherwise. In this case, $f \circ f$ is the identity if and only if $\lambda = \pm 1$. $\lambda = 1$ gives the identity map, so the only nontrivial solutions are $$ f(z) - z_1 = - (z - z_1) \Longleftrightarrow f(z) = 2z_1 - z $$ and $$ \frac{f(z)-z_1}{f(z)-z_2} = - \frac{z-z_1}{z-z_2} \Longleftrightarrow f(z) = \frac{(z_1+z_2)z - 2z_1z_2}{2z - (z_1+z_2)} $$

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In terms of the coefficients in $$ f(z) \ = \ \frac{az+b}{cz+d} $$ $f \circ f = \text{id}$ is equivalent to solve $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} $$ for some $\lambda \ne 0$. This gives the equations $$ a^2 + bc = bc + d^2 = \lambda \\ (a+d)b = (a+d)c = 0 $$ If $a+d \ne 0$ then $b = c = 0$ and $a^2 = d^2 = \lambda \ne 0$, this gives $f(z) = \pm z$. If $a + d = 0$ then the equations reduce to $a^2 + bc = \lambda \ne 0$.

So $f \circ f = \text{id}$ holds if and only if $$ b = c = 0 \text{ and } a^2 = d^2 \ne 0 $$ or $$ a + d = 0 \text{ and } a^2 + bc \ne 0. $$

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These conditions are not satisfied in your example $f(z) = \frac{z + i}{iz+1}$, and actually $f(f(z)) = \frac 1z$ in this case.

Answer 4

This is a partial answer (but is too long for a comment).

All non-empty, non-full, simply connected open sets of the complex plane are conformally isomorphic. Hence, working with the Mobius transformations of the disk $\{z: \ |z| < 1\}$ is, up to conjugation, the same as working with the group of conformal automorphisms of the upper half-plane $\mathbb{H} := \{z: \ Im(z) > 0\}$.

The group of conformal automorphisms of $\mathbb{H}$ is well-known: it is isomorphic to the group $PSL_2 (\mathbb{R})$, which acts by homographies:

$$\left(\begin{matrix}a & b \\ c & d \end{matrix}\right) \cdot z = \frac{az+b}{cz+d}.$$

There is a classification of these automorphisms (the identity excluded) as $\textit{hyperbolic}$, $\textit{parabolic}$ and $\textit{elliptic}$. An automorphism $M$ is hyperbolic if it is conjugated to a diagonal matrix:

$$M \sim \left(\begin{matrix}e^{t/2} & 0 \\ 0 & e^{-t/2}\end{matrix}\right).$$

An automorphism $M$ is parabolic if it is conjugated to a transvection matrix:

$$M \sim \left(\begin{matrix}1 & t \\ 0 & 1\end{matrix}\right).$$

Finally, an automorphism $M$ is parabolic if it is conjugated to a rotation matrix:

$$M \sim \left(\begin{matrix}\cos(t) & sin(t) \\ -\sin(t) & \cos(t)\end{matrix}\right).$$

There is a very nice characterization of these families. Since conjugation preserves the trace, we see that $M$ is:

• hyperbolic if and only if $|Tr(M)| > 2$ (in which case the parameter $t$ is given by $2 Argch (|Tr(M)|/2)$);

• parabolic if and only if $|Tr(M)| = 2$;

• elliptic if and only if $|Tr(M)| < 2$ (in which case $\cos(t) = |Tr(M)|/2$).

Now, what does this has to do with your question? Well, a hyperbolic automorphism acts as a non-trivial translation along some geodesic (an Euclidean half-circle). It cannot be a torsion element (this can be seen easily from the normal form). A parabolic automorphism acts as a non-trivial translation along some horocycle (an Euclidean circle). It cannot be a torsion element (this also can be seen easily from the normal form). An elliptic automorphism is a torsion element if and only if its angle is a rational multiple of $\pi$.

Now, you want the transformation to be its own inverse, that is, $M^2 = \pm I$ (we are working in $PSL_2 (\mathbb{R})$, not $SL_2 (\mathbb{R})$! The matrices $M$ and $-M$ correspond to the same homography.). If the rotation is of angle $t \in [0, \pi)$, then we want $2t \in \{0, \pi\}$, so $t \in \{0, \pi/2\}$. That is, either $M = \pm I$, or $\cos(t) = 0$, so the homography corresponding to a matrix $M$ is an involution if and only if $M = I$ or $Tr(M) = 0$. Finally, we get a two-real-parameters family of solutions:

$$f(z) = \frac{az+b}{cz-a},$$

with [$a^2+bc = -1$ and $a > 0$] or [$bc = -1$ and $b>0$], plus the identity. This can actually be seen as a one-complex-parameter family, each of these matrices doing a rotation by an angle of $\pi$ around some point $z_0 \in \mathbb{H}$, which is the parameter.