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 Incoming links: Conjugation of a Möbius transformation with two distinct fixed points

Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 1 / 8A / ii / Solution Created 2026-09-24 Updated 2026-10-14

For the given Möbius transformation, the fixed point equation is

z (z + 3) = 3 z + 1

, hence

z = 1, z = - 1.

For the general Möbius transformation, its finite fixed points satisfy

c z^{2} + (d - a) z - b = 0.

With

c \neq = 0

and nonzero discriminant, the quadratic formula gives the two distinct fixed points of a Möbius transformation:

α = \frac{a - d + m}{2 c}, β = \frac{a - d - m}{2 c}, m^{2} = (a - d)^{2} + 4 b c .

The fixed-point equation for

α

also gives

b - d α = - α (a - c α)

. Therefore

ω - α = \frac{( a - c α ) ( z - α )}{cz + d}, ω - β = \frac{( a - c β ) ( z - β )}{cz + d} .

Dividing yields the conjugation of a Möbius transformation with two distinct fixed points:

\frac{ω - α}{ω - β} = k \frac{z - α}{z - β}, k = \frac{a - c α}{a - c β} = \frac{a + d - m}{a + d + m} .

Neither factor in this quotient vanishes, since

(a - c α) (a - c β) = a d - b c \neq = 0

. The formula extends over the Riemann sphere, including the transformation's pole. For the special transformation and the ordering

α = 1, β = - 1

,

k = 1/2

.

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