Chapter 9.5, Problem 39E

a.

To determine

To find:First five iterates

Answer to Problem 39E

The five iterates are $2i+1$ , $-2+i$ , $-2i-1$ , $2-i$ & $2i+1$ .

Explanation of Solution

Given:

  $f(z)=iz,z_0=2-i$

Concept Used:

A discrete dynamical system models the evolution of state variables of the system over discrete time steps. We imagine that we take a snapshot of the state variables at a sequence of times. The dynamical system model describes a rule for how to evolve from one snapshot to the next.

Calculation:

Given function & initial value

  $f(z)=iz,z_0=2-i$

  $f(z)=iz$

Multiply previous one with $i$

  $\begin{array}{l}{z}_0=2-i\\ \Rightarrow z_1=i(2-i)=2i-i^2=2i+1\\ \Rightarrow z_2=i(2i+1)=2i^2+i=-2+i\\ \Rightarrow z_3=i(-2+i)=-2i+i^2=-2i-1\\ \Rightarrow z_4=i(-2i-1)=-2i^2-i=2-i\\ \Rightarrow z_5=i(2-i)=2i-i^2=2i+1\end{array}$

Hence,

The five iterates are $2i+1$ , $-2+i$ , $-2i-1$ , $2-i$ & $2i+1$ .

b.

To determine

To find: First five iterates

Answer to Problem 39E

The five iterates are $0.5-0.866i$ , $-0.5-0.866i$ , $-1-0i$ , $-0.5+0.866i$ & $0.5+0.866i$ .

Explanation of Solution

Given:

  $f(z)=(0.5-0.866i)z,z_0=1+0i$

Concept Used:

A discrete dynamical system models the evolution of state variables of the system over discrete time steps. We imagine that we take a snapshot of the state variables at a sequence of times. The dynamical system model describes a rule for how to evolve from one snapshot to the next.

Calculation:

Given function & initial value

  $f(z)=(0.5-0.866i)z,z_0=1+0i$

  $f(z)=(0.5-0.866i)z$

Multiply previous one with $(0.5-0.866i)$

  $\begin{array}{l}{z}_0=1+0t\\ \Rightarrow z_1=(0.5-0.866i)(1+0t)=0.5-0.866i\\ \Rightarrow z_2=(0.5-0.866i)(0.5-0.866i)=0.25-0.866i-0.75=-0.5-0.866i\\ \Rightarrow z_3=(0.5-0.866i)(-0.5-0.866i)=-0.25-0.75=-1-0i\\ \Rightarrow z_4=(0.5-0.866i)(-1-0i)=-0.5+0.866i\\ \Rightarrow z_5=(0.5-0.866i)(-0.5+0.866i)=-0.25+0.866i+0.75=0.5+0.866i\end{array}$

Hence,

The five iterates are $0.5-0.866i$ , $-0.5-0.866i$ , $-1-0i$ , $-0.5+0.866i$ & $0.5+0.866i$ .