Chapter 9.8, Problem 11CFU

To determine

To find whether the given complex number is in prisoner set or escape set.

Answer to Problem 11CFU

Escape set.

Explanation of Solution

Given:

The complex number is $w=0.8-07i$ for the function $f(z)=z^2$ .

Calculation:

Determine $r=\sqrt{x^2+y^2}$

  $\begin{array}{l}r=\sqrt{{\left(0.8\right)}^2+{\left(-0.7\right)}^2}\\ r=\sqrt{1.13}\\ r=1.06\end{array}$

Now, determine $\theta =\arctan \left(\frac{y}{x}\right)$

  $\begin{array}{l}\theta =\arctan \left(\frac{-0.7}{0.8}\right)\\ \theta =\arctan \left(-0.875\right)\\ \theta =-0.72\end{array}$

The polar form of $w=w=0.8-07i=1.06\left[\cos (-0.72)+i\sin (-0.72)\right]$

Now, iterate the function

  $\begin{array}{l}{w}_1=f(w)\\ =w^2\\ ={\left(1.06\left[\cos (-0.72)+i\sin (-0.72)\right]\right)}^2\end{array}$

And

  $\begin{array}{l}{w}_2=f(w_1)\\ =w^4\\ ={\left(1.06\left[\cos (-0.72)+i\sin (-0.72)\right]\right)}^4\end{array}$

And so on...........

The moduli of these iterate are ${(1.06)}^2,{(1.06)}^4,{(1.06)}^8\mathrm{.............}$ and so on. These moduli will approach to infinite $\infty$ as the number of iterations increase. This means the graph of iterates escape to infinity in the complex plane.

So, $w=0.8-07i$ is in the escape set of the function.