(i) State de Moivre's theorem. Use it to express $\cos 5 \theta$ as a polynomial in $\cos \theta$.

(ii) Find the two fixed points of the Möbius transformation

$$z \longmapsto \omega=\frac{3 z+1}{z+3}$$

that is, find the two values of $z$ for which $\omega=z$.

Given that $c \neq 0$ and $(a-d)^{2}+4 b c \neq 0$, show that a general Möbius transformation

$$z \longmapsto \omega=\frac{a z+b}{c z+d}, \quad a d-b c \neq 0,$$

has two fixed points $\alpha, \beta$ given by

$$\alpha=\frac{a-d+m}{2 c}, \quad \beta=\frac{a-d-m}{2 c}$$

where $\pm m$ are the square roots of $(a-d)^{2}+4 b c$.

Show that such a transformation can be expressed in the form

$$\frac{\omega-\alpha}{\omega-\beta}=k \frac{z-\alpha}{z-\beta},$$

where $k$ is a constant that you should determine.