Let $T(z)=\frac{a_1z + b_1}{c_1z+d_1}$ $S(z)=\frac{a_2z+b_2}{c_2z+d_2}$ be two Mobius transformations. Show that if …

Question

Let $T(z)=\frac{a_1z + b_1}{c_1z+d_1}$ $S(z)=\frac{a_2z+b_2}{c_2z+d_2}$ be two Mobius transformations. Show that if $T = S$, then there is $\lambda \in \mathbb C$ such that $a_1 = \lambda a_2, b_1= \lambda b_2, c_1 = \lambda c_2, d_1 = \lambda d_2$.

$\frac{a_1z + b_1}{c_1z+d_1} \leftrightarrow c_1z^2 +(d_1 – a_1)z + b_1$

$\frac{a_2z+b_2}{c_2z+d_2} \leftrightarrow c_2z^2 +(d_2 – a_2)z + b_2 = 0$

$c_1z^2 +(d_1 – a_1)z + b_1 = c_2z^2 +(d_2 – a_2)z + b_2 = 0$

$c_1 = 1 c_2 $

$b_1 = 1 b_2 $

$d_1 – a_1 = d_2 – a_2 \to a_1 = d_1 = 1 a_2 = 1 d_2$.

This must not be right because it's too simple, but it was the only idea I had.

Thank's for any help.

Answer 1

(Assuming $a_1, a_2, b_1, b_2, c_1, c_2, d_1, d_2$ all non-null to keep the answer short: these have to be dealt with in a few separate cases).

You can use specific values for $z$:

• $z=0$ gives $b_1/d_1=b_2/d_2$
• $z\rightarrow +\infty$ gives $a_1/c_1=a_2/c_2$
• $z=-b_1/a_1$ (zero) gives $a_2/a_1=b_2/b_1$
• (as an alternative to any of the 3 previous lines) $z\rightarrow -d_1/c_1$ (pole) gives $c_2/c_1=d_2/d_1$

So posing $a_1/a_2=\lambda$, we get
$\lambda=c_1/c_2=b_1/b_2=d_1/d_2$