TITLE: When is a mobius transformation its own inverse?
QUESTION [6 upvotes]: 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:

REPLY [7 votes]: 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 $c=0$.