Fixed Points

Let $m=\begin{bmatrix}a & b \\ c& d\end{bmatrix}$ be a matrix of determinant 1. The fixed points of $m(z)$ satisfy $m(z)=z$. This is a quadratic equation $$cz^2+(d-a) z -b =0.$$ Thus, the transformation has one or two fixed points. The type of the transformation is as follows

Parabolic:

Precisely one fixed point. This requires the discriminant to vanish, which means $(d-a)^2+4bc= (a+d)^2 -4=0$. Hence the Trace of $m$ must be $a+d=\pm2$.

Hyperbolic:

Two distinct fixed points, one of which is attractive, and one of which is repelling. This requires $\mathop{\text{Trace}}m \notin [-2,2]$.

Elliptic:

Two distinct fixed points, both of which are neutral. The trace lies in the interval $(-2,2)$.

Just for completeness, let's consider the fixed points of the anticonformal transformation. This is the equation $m(\bar z) = z$, which is equivalent to
$$\begin{align*} 0&=z(c\bar z+d) -a\bar z - b\\ &= c z\bar z +d z - a \bar z -b... ...trix}c & -a \\ d & -b\end{bmatrix} \begin{bmatrix}z\\ 1\end{bmatrix}\end{align*}$$
This looks like the equation of a circle, as we explained in our notes on circles and Hermitian matrices, except that the matrix $\begin{bmatrix}c & -a \\ d & -b\end{bmatrix}$ in the middle may not be Hermitian. However, if we take the conjugate transpose of that equation, we get an equivalent equation $$0 = \begin{bmatrix}\bar z & 1\end{bmatrix} \begin{bmatrix}\b... ...bar a & -\bar b\end{bmatrix}\begin{bmatrix}z\\ 1\end{bmatrix}.$$ If we add these two valid equations together, we get another valid equation $$0 = \begin{bmatrix}\bar z & 1\end{bmatrix} \begin{bmatrix}c+... ...bar a & -\bar b-b\end{bmatrix}\begin{bmatrix}z\\ 1\end{bmatrix}$$ and the matrix $$H= \begin{bmatrix}\mathop{\text{Re}}c & \bar d -a \\ d -\bar a & -\mathop{\text{Re}}b\end{bmatrix}$$ is indeed Hermitian. If the determinant is negative, then we know that the fixed points form a whole circle. The determinant is $$-\vert\bar d -a\vert^2 -\mathop{\text{Re}}c \mathop{\text{Re}}b.$$ If the determinant is positive, there are no solutions and hence no fixed points. If the determinant is 0, there are usually also no solutions, unless the whole matrix $H$ is 0.