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Two exercises on complex sequences (one about Mandelbrot set)

  1. Jan 1, 2014 #1
    The problem statement, all variables and given/known data.

    I am trying to solve two exercises about complex sequences:

    1) Let ##\alpha \in \mathbb C##, ##|\alpha|<1##. Which is the limit ##\lim_{n \to \infty} \alpha^n##?, do the same for the case ##|\alpha|>1##.

    2) Let ##\mathcal M## be the set of the complex numbers ##c## such that the sequence defined recursively by: ##z_0=0##, ##z_{n+1}={z_n}^2+c## is bounded (##\mathcal M## is called the Mandelbrot set). Prove that ##\mathcal M \subset \{|z|\leq 2\}##

    The attempt at a solution.

    For point 1) what I did was: if ##|\alpha|<1##, I can express ##\alpha## in its exponential form, so ##\alpha=re^{iθ}## for ##0<r<1## and ##0\leq θ<2\pi##. Then, ##\alpha^n=r^ne^{niθ}##. Lets show that ##\alpha^n \to 0##, ##0\leq |r^ne^{niθ}|=|r^n(\cos(θ)+i\sin(θ))|\leq 2|r|^n \to 0##.

    I don't know what to do for the case ##|\alpha|>1##. In this case, if I consider the vector from the point ##(0,0)## to ##(a,b)##, where ##\alpha^n=a+ib##, then the lenght of the vector tends to infinity. But I don't what to say about the limit of ##\alpha^n##. From my observation, I could conclude that ##|\alpha^n| \to \infty##. Would this imply that the sequence ##\{\alpha^n\}_{n \in \mathbb N}## doesn't have a limit?.

    For point 2), I don't have any idea how to prove this. Suppose ##\mathcal M \not \subset \{|z|\leq 2\}##. Then, there is some ##c \in \mathcal M : |c|>2##. How could I prove that the sequence ##\{z_n\}_{n \in \mathbb N}## is not bounded?
     
  2. jcsd
  3. Jan 1, 2014 #2
    Yes.

    Note however that ##|r^n e^{ni\theta}| = r^n##. This is a better approximation. But your method was good too.

    Yes.

    Take ##|c|>2##. Maybe you could start with writing out the first 5 or 6 terms of the ##z_n## sequence to see if you can find some pattern. Of course you only care about the norm of ##z_n##, since you need to show that goes to infinity.
     
  4. Jan 1, 2014 #3
    Thanks for the first remark. Now, for the second part, I did write the first terms of the sequence but I can't arrive to an expression that helps me to prove what I want to. I mean, I would like to express ##|z_n|=z^n|c|## with ##z>1## or get to a similar expression in order to conclude that the norm tends to infinity. Could you give me another hint for that?
     
  5. Jan 1, 2014 #4
    You have

    [tex]|z_0| = 0[/tex]
    [tex]|z_1| = |c|[/tex]
    [tex]|z_2| = |z_1^2 + c| \ |c| |c+1|\geq |c|[/tex]
    [tex]|z_3| = |(c^2 + c)^2 + c| = |c^2 (c +1)^2 + c| = |c| |c(c+1)^2 + 1|\geq |c|( |c|^2 |c+1|^2 - 1) \geq |c| (|c|^2 -1) \geq |c|^2[/tex]

    where I make fundamental use of the inequality ##|a-b| \geq ||a| - |b||##.
     
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