Chapter 12.8, Problem 26E

To determine

To find: the first three iterates of the function using the given initial value.

Answer to Problem 26E

  $4+2i,15+18i\text{ and }-96+542i$

Explanation of Solution

Given information:

  $\begin{array}{l}f\left(z\right)=z^2+c\\ c=3+2i,z_0=1\end{array}$

Calculation:

The function would be evaluated as follows:

  $\begin{array}{c}f\left(z\right)=z^2+c\\ =z^2+3+2i\end{array}$

The first three iterates of function using the given initial value can be evaluated as shown below.

  $\begin{array}{c}{z}_1=f\left(z_0\right)\\ ={\left(1\right)}^2+3+2i\left[\text{Put }{z}_0=1\right]\\ =4+2i\\ z_2=f\left(z_1\right)\\ ={\left(4+2i\right)}^2+3+2i\left[\text{Put }{z}_1=4+2i\right]\\ =16+16i-4+3-2i\\ =15+18i\\ z_3=f\left(z_2\right)\\ ={\left(15+18i\right)}^2+2+3i\left[\text{Put }{z}_2=15+18i\right]\\ =225+540i-324+3+2i\\ =-96+542i\end{array}$

Thus, the three iterates of $f\left(z\right)$ are $4+2i,15+18i\text{ and }-96+542i$ .