How can a general automorphism of the unit disc be parabolic or hyperbolic? I am trying to work through Needham's Visual Complex Analysis on my own. We are told that the general automorphism of the unit disc is given by the Mobius transformation $M_a^{\phi}(z)=e^{i\phi}\left(\frac{z-a}{\bar{a}z-1}\right)$.
We are also told that it can be regarded as a composition of the Mobius transformation $M_a(z)=\frac{z-a}{\bar{a}z-1}$ and a rotation through the angle $\phi$.
My problem is that $M(z)=\frac{z-a}{\bar{a}z-1}$ gives an elliptic transformation (as we are told from a table that lists the type of transformation against the value of $|a+d|$). For a transformation to be hyperbolic, $|a+d|$ has to be greater than 2 and $|a+d|$ is zero.
Can the $e^{i\phi}$ multiply the top line of $\frac{z-a}{\bar{a}z-1}$ so that $|a+d|$ is no longer zero? Remembering that the matrix has to be normalized (and that since $|a|\lt1$,  $ad-bc = |a|^2-1 \lt1 $ that dividing by its square root to normalize the matrix will increase |a+d|); is that what one must do? But this method goes against the idea of a composition of two transformations.
I think there is a lot here I have failed to understand but if anyone could see through what may well be muddle (I've done my best, please don't just subtract points!) and give me some clues I would be very grateful.
 A: The transformation $z\mapsto\frac{z-a}{\overline{a}z-1}$ is represented by $M=[\begin{smallmatrix}1&-a\\\overline{a}&-1\end{smallmatrix}]$, but this matrix doesn't have determinant $1$ so it is not suitable for the $\mathrm{tr}^2<4$ test for ellipticity. Instead it must be normalized to the matrix
$$ N=\frac{i}{\sqrt{1-|a|^2}}\begin{bmatrix}1&-a\\\overline{a}&-1\end{bmatrix}, $$
which of course has trace $0$ just like $M$ does.
The transformation $z\mapsto e^{i\phi}z$ may be represented by $S=[\begin{smallmatrix}e^{i\phi}&0\\0&1\end{smallmatrix}]$. We can multiply $SM$ out to get a matrix which represents $e^{i\phi}M_a(z)$, and the effect will be to multiply the top row of $M$ by $e^{i\phi}$. We could do the same for the product $SN$. But in general neither will have determinant $1$, since $S$ doesn't have determinant $1$, and so this is unsuitable for the $\mathrm{tr}^2$ test again. Instead, we must normalize $S$ to $R=[\begin{smallmatrix}e^{i\phi/2}&0\\0&e^{-i\phi/2}\end{smallmatrix}]$ and compute the product $RN$. The effect is to multiply the top row of $N$ by $e^{i\phi/2}$ and the bottom row by $e^{-i\phi/2}$.
The result is a matrix $RN$ which represents $e^{i\phi}M_a(z)$, has determinant $1$, and has trace
$$ \mathrm{tr}(RN)=-\frac{2\sin(\phi/2)}{\sqrt{1-|a|^2}}, $$
can indeed satisfy $\mathrm{tr}^2\ge4$ thus $e^{i\phi}M_a(z)$ can be parabolic or hyperbolic if $e^{i\phi}$ is different enough from $1$. What you've just encountered, then, is the fact that the composition of two elliptic transformations ($e^{i\phi}z$ and $M_a(z)$) may not itself be elliptic.
Is this actually surprising? Elliptic transformations are supposed to behave like rotations of the Riemann sphere, and rotations are of course closed under composition, so from that description alone it may be surprising. But only elements of $\mathrm{SU}(2)$ (which intersects $\mathrm{Aut}\,\mathbb{D}=\mathrm{SU}(1,1)$ precisely at $S(U(1)^2)$, the usual rotations of $\mathbb{C}$) act as rotations of the Riemann sphere with the usual round Fubini-Study metric (inherited from identifying $\widehat{\mathbb{C}}$ with $\mathbb{CP}^1$ and $\mathbb{CP}^1$ with projectors in the space of $2\times2$ complex hermitian matrices under the Hilbert-Schmidt norm / Frobenius inner product). If you conjugate a rotation (WRT the usual metric) by a non-rotation, the result is something that looks like a rotation but WRT a different metric. If two transformations look like rotations but WRT different metrics, why should we expect the composition to look like a rotation at all?
We can look at a particular example as a sanity check. Set $e^{i\phi}=-1$ and $a=1/2$ so the Mobius transformation is $\frac{~\,2z-1}{-z\,+\,2~}$. The transformation $e^{i\phi}z=-z$ is obviously elliptic, it rotates $180^\circ$ around the origin. To understand the transformation $M_a$, imagine we picked appropriate points $p\in(0,1),q\in(1,\infty)$ and created bipolar coordinates using them; then $M_a$ would be a $180^\circ$ "bipolar" rotation around both poles, which restricts to a $180^\circ$ rotation of the unit circle as well as a $180^\circ$ rotation of the circle containing a diameter between $0$ and $a$ (thus swapping $0$ and $a$). This is also clearly elliptic. The composition must then fix $\pm1$, and if we consider the effect of it on any point in $(-1,1)$ we see it moves closer to $-1$ from $+1$. The same is true for all arcs between $\pm1$ that fill up the unit circle, so the composition is indeed hyperbolic.
