Chapter 1.6, Problem 76PS
a.
To State:
Whether the given number z0 belongs to the Julia set associated with the function fz=z2+1+i .
Since z1=i=z0 , application of f on z0=i gives i itself, it is a fixed point of f . The value of zn remains finite as n→∞ and hence, z0=i belongs to the Julia Set for c=1+i .
Given:
z0=i
Concepts Used:
For fz=z2+c , if the value of zn has an upper bound for all natural numbers n , then z0 is included in the Julia set for c . Here, zn represents the result of applying f on z0 for n times successively, for example z5=fffffz0 .
A value x=a is known as a fixed point of a function gx if ga=a .
Calculations:
Calculate z1=fz0 for fz=z2+1+i and z0=i .
z1=fz0z1=fiz1=i2+1+iz1=−1+1+iz1=i=z0
Conclusion:
Since z1=i=z0 , application of f on z0=i gives i itself, it is a fixed point of f . The value of zn remains finite as n→∞ and hence, z0=i belongs to the Julia Set for c=1+i .
b.
To State:
Whether the given number z0 belongs to the Julia set associated with the function fz=z2+1+i .
Since z1≈2.236 is greater than R≈1.79 , z0=1 does not belong to the Julia Set for c=1+i .
Given:
z0=1
Concepts Used:
For fz=z2+c and R=1+1+4c2 , if for some natural number n>0 and complex number z0 we have zn>R , then z0 is not included in the Julia set for c . Here, zn represents the result of applying f on z0 for n times successively, for example z5=fffffz0 .
Calculations:
Calculate z1=fz0 for fz=z2+1+i and z0=1 .
z1=fz0z1=f1z1=12+1+iz1=2+i
Since z1≠z0 , compare their moduli.
Calculate z0 :
z0=1=1
Calculate z1 :
z1=2+i=22+12=5≈2.236
Observe that z1>z0 .
Calculate the limiting value R=1+1+4c2 , here c=1+i .
R=1+1+4c2R=1+1+41+i2R=1+1+412+122R=1+1+422R≈1.79
Note that z1>R .
Conclusion:
Since z1≈2.236 is greater than R≈1.79 , z0=1 does not belong to the Julia Set for c=1+i .
c.
To State:
Whether the given number z0 belongs to the Julia set associated with the function fz=z2+1+i .
Since z1≈3.16 is greater than R≈1.79 , z0=2i does not belong to the Julia Set for c=1+i .
Given:
z0=2i
Known from previous part:
The limiting value R=1+1+4c2 for the present case where c=1+i is ≈1.79 .
Concepts Used:
For fz=z2+c and R=1+1+4c2 , if for some natural number n>0 and complex number z0 we have zn>R , then z0 is not included in the Julia set for c . Here, zn represents the result of applying f on z0 for n times successively, for example z5=fffffz0 .
Calculations:
Calculate z1=fz0 for fz=z2+1+i and z0=2i .
z1=fz0z1=f2iz1=2i2+1+iz1=−4+1+iz1=−3+i
Note that z1≠z0
Calculate z1 :
z1=−3+i=−32+12=10≈3.16
Note that z1>R .
Conclusion:
Since z1≈3.16 is greater than R≈1.79 , z0=2i does not belong to the Julia Set for c=1+i .
d.
To State:
Whether the given number z0 belongs to the Julia set associated with the function fz=z2+1+i .
Since z1≈13.60 is greater than R≈1.79 , z0=2+3i does not belong to the Julia Set for c=1+i .
Given:
z0=2+3i
Known from previous part:
The limiting value R=1+1+4c2 for the present case where c=1+i is ≈1.79 .
Concepts Used:
For fz=z2+c and R=1+1+4c2 , if for some natural number n>0 and complex number z0 we have zn>R , then z0 is not included in the Julia set for c . Here, zn represents the result of applying f on z0 for n times successively, for example z5=fffffz0 .
Calculations:
Calculate z1=fz0 for fz=z2+1+i and z0=2+3i .
z1=fz0z1=f2+3iz1=2+3i2+1+iz1=4−9+12i+1+iz1=−4+13i
Note that z1≠z0
Calculate z1 :
z1=−4+13i=−42+132=16+169=185≈13.60
Note that z1>R .
Conclusion:
Since z1≈13.60 is greater than R≈1.79 , z0=2+3i does not belong to the Julia Set for c=1+i .