Chapter 1.6, Problem 76PS

Solution

a.

To State:

Whether the given number $z_0$ belongs to the Julia set associated with the function $f\left(z\right)=z^2+1+i$ .

Since $z_1=i=z_0$ , application of $f$ on $z_0=i$ gives $i$ itself, it is a fixed point of $f$ . The value of $\left|z_n\right|$ remains finite as $n\to \infty$ and hence, $z_0=i$ belongs to the Julia Set for $c=1+i$ .

Given:

  $z_0=i$

Concepts Used:

For $f\left(z\right)=z^2+c$ , if the value of $\left|z_n\right|$ has an upper bound for all natural numbers $n$ , then $z_0$ is included in the Julia set for $c$ . Here, $z_n$ represents the result of applying $f$ on $z_0$ for $n$ times successively, for example $z_5=f\left(f\left(f\left(f\left(f\left(z_0\right)\right)\right)\right)\right)$ .

A value $x=a$ is known as a fixed point of a function $g\left(x\right)$ if $g\left(a\right)=a$ .

Calculations:

Calculate $z_1=f\left(z_0\right)$ for $f\left(z\right)=z^2+1+i$ and $z_0=i$ .

  $\begin{array}{l}{z}_1=f\left(z_0\right)\\ z_1=f\left(i\right)\\ z_1=i^2+1+i\\ z_1=-1+1+i\\ z_1=i=z_0\end{array}$

Conclusion:

Since $z_1=i=z_0$ , application of $f$ on $z_0=i$ gives $i$ itself, it is a fixed point of $f$ . The value of $\left|z_n\right|$ remains finite as $n\to \infty$ and hence, $z_0=i$ belongs to the Julia Set for $c=1+i$ .

b.

To State:

Whether the given number $z_0$ belongs to the Julia set associated with the function $f\left(z\right)=z^2+1+i$ .

Since $\left|z_1\right|\approx 2.236$ is greater than $R\approx 1.79$ , $z_0=1$ does not belong to the Julia Set for $c=1+i$ .

Given:

  $z_0=1$

Concepts Used:

For $f\left(z\right)=z^2+c$ and $R=\frac{1+\sqrt{1+4\left|c\right|}}{2}$ , if for some natural number $n>0$ and complex number $z_0$ we have $\left|z_n\right|>R$ , then $z_0$ is not included in the Julia set for $c$ . Here, $z_n$ represents the result of applying $f$ on $z_0$ for $n$ times successively, for example $z_5=f\left(f\left(f\left(f\left(f\left(z_0\right)\right)\right)\right)\right)$ .

Calculations:

Calculate $z_1=f\left(z_0\right)$ for $f\left(z\right)=z^2+1+i$ and $z_0=1$ .

  $\begin{array}{l}{z}_1=f\left(z_0\right)\\ z_1=f\left(1\right)\\ z_1=1^2+1+i\\ z_1=2+i\end{array}$

Since $z_1\ne z_0$ , compare their moduli.

Calculate $\left|z_0\right|$ :

  $\begin{array}{c}\left|z_0\right|=\left|1\right|\\ =1\end{array}$

Calculate $\left|z_1\right|$ :

  $\begin{array}{c}\left|z_1\right|=\left|2+i\right|\\ =\sqrt{2^2+1^2}\\ =\sqrt{5}\\ \approx 2.236\end{array}$

Observe that $\left|z_1\right|>\left|z_0\right|$ .

Calculate the limiting value $R=\frac{1+\sqrt{1+4\left|c\right|}}{2}$ , here $c=1+i$ .

  $\begin{array}{l}R=\frac{1+\sqrt{1+4\left|c\right|}}{2}\\ R=\frac{1+\sqrt{1+4\left|1+i\right|}}{2}\\ R=\frac{1+\sqrt{1+4\sqrt{1^2+1^2}}}{2}\\ R=\frac{1+\sqrt{1+4\sqrt{2}}}{2}\\ R\approx 1.79\end{array}$

Note that $\left|z_1\right|>R$ .

Conclusion:

Since $\left|z_1\right|\approx 2.236$ is greater than $R\approx 1.79$ , $z_0=1$ does not belong to the Julia Set for $c=1+i$ .

c.

To State:

Whether the given number $z_0$ belongs to the Julia set associated with the function $f\left(z\right)=z^2+1+i$ .

Since $\left|z_1\right|\approx 3.16$ is greater than $R\approx 1.79$ , $z_0=2i$ does not belong to the Julia Set for $c=1+i$ .

Given:

  $z_0=2i$

Known from previous part:

The limiting value $R=\frac{1+\sqrt{1+4\left|c\right|}}{2}$ for the present case where $c=1+i$ is $\approx 1.79$ .

Concepts Used:

For $f\left(z\right)=z^2+c$ and $R=\frac{1+\sqrt{1+4\left|c\right|}}{2}$ , if for some natural number $n>0$ and complex number $z_0$ we have $\left|z_n\right|>R$ , then $z_0$ is not included in the Julia set for $c$ . Here, $z_n$ represents the result of applying $f$ on $z_0$ for $n$ times successively, for example $z_5=f\left(f\left(f\left(f\left(f\left(z_0\right)\right)\right)\right)\right)$ .

Calculations:

Calculate $z_1=f\left(z_0\right)$ for $f\left(z\right)=z^2+1+i$ and $z_0=2i$ .

  $\begin{array}{l}{z}_1=f\left(z_0\right)\\ z_1=f\left(2i\right)\\ z_1={\left(2i\right)}^2+1+i\\ z_1=-4+1+i\\ z_1=-3+i\end{array}$

Note that $z_1\ne z_0$

Calculate $\left|z_1\right|$ :

  $\begin{array}{c}\left|z_1\right|=\left|-3+i\right|\\ =\sqrt{{\left(-3\right)}^2+1^2}\\ =\sqrt{10}\\ \approx 3.16\end{array}$

Note that $\left|z_1\right|>R$ .

Conclusion:

Since $\left|z_1\right|\approx 3.16$ is greater than $R\approx 1.79$ , $z_0=2i$ does not belong to the Julia Set for $c=1+i$ .

d.

To State:

Whether the given number $z_0$ belongs to the Julia set associated with the function $f\left(z\right)=z^2+1+i$ .

Since $\left|z_1\right|\approx 13.60$ is greater than $R\approx 1.79$ , $z_0=2+3i$ does not belong to the Julia Set for $c=1+i$ .

Given:

  $z_0=2+3i$

Known from previous part:

The limiting value $R=\frac{1+\sqrt{1+4\left|c\right|}}{2}$ for the present case where $c=1+i$ is $\approx 1.79$ .

Concepts Used:

For $f\left(z\right)=z^2+c$ and $R=\frac{1+\sqrt{1+4\left|c\right|}}{2}$ , if for some natural number $n>0$ and complex number $z_0$ we have $\left|z_n\right|>R$ , then $z_0$ is not included in the Julia set for $c$ . Here, $z_n$ represents the result of applying $f$ on $z_0$ for $n$ times successively, for example $z_5=f\left(f\left(f\left(f\left(f\left(z_0\right)\right)\right)\right)\right)$ .

Calculations:

Calculate $z_1=f\left(z_0\right)$ for $f\left(z\right)=z^2+1+i$ and $z_0=2+3i$ .

  $\begin{array}{l}{z}_1=f\left(z_0\right)\\ z_1=f\left(2+3i\right)\\ z_1={\left(2+3i\right)}^2+1+i\\ z_1=4-9+12i+1+i\\ z_1=-4+13i\end{array}$

Note that $z_1\ne z_0$

Calculate $\left|z_1\right|$ :

  $\begin{array}{c}\left|z_1\right|=\left|-4+13i\right|\\ =\sqrt{{\left(-4\right)}^2+{13}^2}\\ =\sqrt{16+169}\\ =\sqrt{185}\\ \approx 13.60\end{array}$

Note that $\left|z_1\right|>R$ .

Conclusion:

Since $\left|z_1\right|\approx 13.60$ is greater than $R\approx 1.79$ , $z_0=2+3i$ does not belong to the Julia Set for $c=1+i$ .