#1

Suppose we iterate the function f(z)=2z^2-i from an initial value z_0 satisfying |z_0|>1.

• Show that the orbit of z_0 under iteration of f diverges to \infty.
• Show that the critical orbit divergest to \infty.

#2

I know we talked about this in class, but can somebody who understood it a bit better than I did type up an explanation? I have a really hard time understanding things if I only hear them.

I understand that the orbit of z_0 under iteration of f does diverge to \infty. I suspect that it has something to do with how z_0 needs to be less than z_0^2.

Where does the importance of the -i come from? It looks to me like the -i doesn’t have much of a say in whether the thing diverges or not; the 2z^2 part is going to have a much larger impact on the next term. Whatever number that is will be at least double the magnitude of -i for z_0>1. Right?

#3

The way that I think about it, is that if z_0>1, then z_0^2 >1 as well. Then it follows that 2z_0^2>2. So then, |2z_0^2-i|>\sqrt{5}, since if z_0=1, the distance would be \sqrt{5}. After one iteration, |2z_1^2-i| > |2(2-i)^2-i|=3\sqrt{13}. After our third iteration, the distance would be \sqrt{55,189}. Clearly we’re increasing, and our rate of change is positive, so heuristically, it makes sense that our orbit would tend towards infinity.

I’m not fully sure if derivatives work the same in \mathbb{C}, but if they do, we can look at the derivative, f'(z_0) = 4z_0, and note that we have a positive rate of change. Thus, we’re increasing more with each step, so we we’ll diverge to infinity.

#4

I don’t think I believe this. If, for example, z_0 = (1+\varepsilon)e^{\pi i/4}, then

2z_0^2 - i = 2(1+\varepsilon)^2 i - i = i + 2\varepsilon + \varepsilon^2 i.

Now, that last expression can be made as close to i as we like but taking \varepsilon to be as close as we need to zero. Thus, we’re certainly not guaranteed that it’s absolute value exceeds \sqrt{5}.

#5

Here’s the simplified version that we came up with in class. First, we’re going to apply the Reverse Triangle Inequality, which states that

|x-y| \geq \left| |x| - |y| \right|.

Applying that here, we get

\begin{align} |f(z_0)| &= |2z_0^2-i| \geq \left| |2z_0^2| - |i| \right| \geq 2|z_0|^2 - 1 \\ &\geq 2|z_0|^2 - |z_0| = |z_0|(2|z_0|-1)= \lambda |z_0|, \end{align}

where \lambda is a number bigger than 1. Since |f(z_0)|>|z_0|>1, we can apply the same logic to obtain

|f^2(z_0)| \geq \lambda^2 |z_0|

and, by induction,

|f^n(z_0)| \geq \lambda^n |z_0|.

As a result, we must have |f^n(z_0)|\to\infty as n\to\infty.