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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.

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2 Answers 2

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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.

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  • $\begingroup$ By following your suggestions, I could construct a transformation $H=S \circ T \circ T^{-1}$ such that $0$ and $\infty$ are the fixed points of $H$ ($T$ has two fixed points). I don't understand why, as you've said, for transformations with two fixed points, it is most convenient if those are $0$ and $\infty$. Could you explain that to me? I suppose that by having those fixed points, it is easy to study the limit for $H$ and from there deduce the limit for $T$. $\endgroup$
    – user156441
    Commented Jun 24, 2014 at 22:21
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    $\begingroup$ A Möbius transformation with $0$ and $\infty$ as the fixed points is of the form $M\colon z \mapsto \alpha\cdot z$. If $\lvert\alpha\rvert < 1$, then iterating $M$ makes the sequence $M^n(z_0)$ converge to $0$ for every $z_0\neq\infty$. If $\lvert\alpha\rvert > 1$, iterating $M$ will let the sequence converge to $\infty$ (unless the starting point is $0$). If $\lvert\alpha\rvert = 1$, then $M$ is a rotation, and $M^n(z)$ converges for no $z$ except $z\in \{0,\infty\}$. $\endgroup$ Commented Jun 24, 2014 at 22:31
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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 $T^n(1) \to -3$.

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  • $\begingroup$ I don't know how can I verify this: $\dfrac{-5}{2}$ is a repelling fixed point and $-3$ is an attractive fixed point. And I also don't understand what it means "close enough", how close enough? Why if a point $z$ is close enought to the attractive fixed point then $T^n(z) \to -3$? Sorry if these questions are too obvious but I really don't get those parts. $\endgroup$
    – user156441
    Commented Jun 24, 2014 at 22:14
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    $\begingroup$ If $T(z_0) = z_0$, then $T(z)- z_0 = T'(z_0)(z-z_0) + O((z-z_0)^2)$, so if $|T'(z_0)| < 1$ and $z$ is close enough to $z_0$ then $|T(z)-z_0| < c|z-z_0|$ for some $c < 1$ (say $c=1+|T'(z_0)|/2$). "Close enough" can be taken as "in a neighbourhood $U$ of $z_0$ so that $|T'(z)| < 1$ on $U$". $\endgroup$
    – mrf
    Commented Jun 24, 2014 at 22:25
  • $\begingroup$ The computations are easier if you conjugate with a suitable Möbius transformation as Daniel suggested. $\endgroup$
    – mrf
    Commented Jun 24, 2014 at 22:27

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