Algebra and Geometry | Part IA, 2006

(i) State de Moivre's theorem. Use it to express cos5θ\cos 5 \theta as a polynomial in cosθ\cos \theta.

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

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

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

Given that c0c \neq 0 and (ad)2+4bc0(a-d)^{2}+4 b c \neq 0, show that a general Möbius transformation

zω=az+bcz+d,adbc0,z \longmapsto \omega=\frac{a z+b}{c z+d}, \quad a d-b c \neq 0,

has two fixed points α,β\alpha, \beta given by

α=ad+m2c,β=adm2c\alpha=\frac{a-d+m}{2 c}, \quad \beta=\frac{a-d-m}{2 c}

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

Show that such a transformation can be expressed in the form

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

where kk is a constant that you should determine.

Typos? Please submit corrections to this page on GitHub.