An archived instance of discourse for discussion in undergraduate Complex Dynamics.

# Nature of a parabolic Mobius transformation

mark

Suppose that a Mobius transformation $T$ has a single fixed point. Show that $T$ is conjugate to some shift $f(z)=z+\alpha$.

Rick

First, I observed that $\infty$ is a fixed point for $f$.

Also, if $T(z)=\frac{az+b}{cz+d}=z$ Then $az+b=cz^2+dz \rightarrow 0=cz^2+(d-a)z-b$.
If $c\neq0$, then the system must have 2 fixed points.

We then examine the case where $c=0$.
By the quadratic equation we can determine that $\infty$ is our fixed point.

Then $0=(d-a)z-b \rightarrow z=\frac{b}{d-a}$ implies that $a=d$ is another condition.

Since $\infty$ is fixed in both cases, we can conclude that $T$ is conjugate to some shift $f(z)=z+\alpha$.

mark

Hmm... I note that
$$T(z) = \frac{2z-1}{z+4}$$
has just one fixed point, even though $c\neq0$. More generally, your quadratic can have a single repeated root.

Nonetheless, you've hit on a crucial point - namely that $\infty$ is a fixed point for $f$ - the only one, if fact.

Rick

Working off of that, i'd suppose that the value under the square root of the quadratic formula must be zero for there to be a single fixed point such that $0=(d-a)^2+4cb$, although this would imply that $z=-\frac{(d-a)}{2c}$. In reference to the above mobius transformation, this would imply a fixed point at $z=-1$. Perhaps we'd want to apply a formula such as $h(z)=\frac{1}{z-\gamma}$, such that a fixed point, $\gamma$, is located at $\infty$? perhaps

Rick

Oh ya, exactly and absolutely