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?
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?
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