Two exercises on complex sequences (one about Mandelbrot set)

[mahler1]

Homework Statement .

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, $\alpha^n=r^ne^{niθ}$. Let's 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 length 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?

 

[R136a1]

Homework Statement .

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, $\alpha^n=r^ne^{niθ}$. Let's show that $\alpha^n \to 0$, $0\leq |r^ne^{niθ}|=|r^n(\cos(θ)+i\sin(θ))|\leq 2|r|^n \to 0$.

Yes.

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

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 length 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?.

Yes.

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?

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.

 

[mahler1]

Yes.

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

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?

 

[R136a1]

You have

|z_0| = 0
|z_1| = |c|
|z_2| = |z_1^2 + c| \ |c| |c+1|\geq |c|
|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

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