Sequence of Mobius Transformation

Question

Let $ T(z) = \frac {z+2}{2z+1} $. Now it follows that: $ T_1(z) = T(z), T_2(z) = T(T_1(z)), T_3(z)=T(T_2(z)) .... T_{n+1}(z)=T(T_n(z)) $ I'm trying to prove this sequence at the nth terms, but I don't see a clear pattern. I know that the $ T(z) = \frac {az+b}{cz+d} $ and basically it follows that pattern: $ \frac {( \alpha + 1)z+ \alpha}{ \alpha z + ( \alpha + 1)} $. Please help with proving this sequence. Thanks

Answer 1

Hint: Use induction to show that

$$ T^{(n)}(z) = \frac{a_n + (a_n + (-1)^{n+1}) z}{(a_n + (-1)^{n+1}) + a_n z}$$

where $$a_n = \frac{3^{n+1} + (-1)^n}{2}$$ Note that I have used a convention starting at $n=0$ so $T^{(0)}(z) = T(z)$.