Limit of complex numbers' sequence (related to Möbius transformation)

Question

Problem Let $T(z)=\dfrac{7z+15}{-2z-4}$. Let $z_1=1$ and $z_n=T(z_{n-1})$ for $n\geq 2$ Find $\lim_{z_n \to \infty}z_n$

I am having a lot of difficulties trying to solve this. I've tried to find a general formula for the sequence but I couldn't. I don't know what properties or theorems to use in order to solve the exercise, I would appreciate any suggestions or hints. Thanks in advance.

Answer 1

You probably know that every Möbius transformation other than the identity has exactly one or two fixed points.

It is much easier to investigate the mapping behaviour for Möbius transformations where the fixed points are in certain standard locations.

For transformations with only one fixed point, it is most convenient if the fixed point is at $\infty$, and for transformations with two fixed points, if the fixed points are located at $0$ and $\infty$.

So find the fixed points of $T$, and conjugate $T$ with a Möbius transformation $S$ mapping the fixed points of $T$ to $0$ and $\infty$ if $T$ has two fixed points, or the fixed point of $T$ to $\infty$ if $T$ has only one fixed point. The behaviour of $S\circ T \circ S^{-1}$ will be obvious.

Answer 2

First of all, the limit must satisfy $T(z) = z$. (Why?)

Your function has two fixpoints, namely $z=-3$ and $z=-5/2$. Furthermore, $|T'(-5/2)| = |2| > 1$, so that fixpoint is repelling. On the other hand, $|T'(-3)| = |1/2| < 1$ which shows that $z=-3$ is an attractive fixpoint. Hence, if $z$ is close enough to $-3$, then $|T(z)-(-3)| < |z-(-3)|$. Finally, some tedious algebra shows that $T(1) = -11/3$ is close enough to $-3$, and hence $T^n(1) \to -3$.