Hello! I'm stuck on how to do this problem and would appreciate some help!
Choose a non-zero complex number c for f(z) = z2+ c. Find a complex number v that is in the escape set. Find a complex number w that is in the prisoner set. Graph the first five iterations of both v and w and connect them with line segments.
Find a complex number j that is in the Julia set.
I chose c=2i. I know that the escape set is any point that escapes to infinity under iteration. The prisoner set is a point that does not escape. The Julia set is the boundary between the escape set and the prisoner set.
I've looked in my textbook and online but nothing has exactly what I need. Thanks in advance for your help!
Sure! I can help you with that.
To find a complex number v that is in the escape set, we need to iterate the function f(z) = z^2 + c multiple times and check if the absolute value of the result goes to infinity. Let's start by selecting a value for v and applying f(z) to it.
Let's take v = 0 as an initial value. Plug this into the function: f(0) = 0^2 + c = c.
Now we need to apply the function iteratively to see if it escapes to infinity. We'll do this five times:
- Iteration 1: f(c) = c^2 + c
- Iteration 2: f(f(c)) = (c^2 + c)^2 + c
- Iteration 3: f(f(f(c))) = ((c^2 + c)^2 + c)^2 + c
- Iteration 4: f(f(f(f(c)))) = (((c^2 + c)^2 + c)^2 + c)^2 + c
- Iteration 5: f(f(f(f(f(c))))) = ((((c^2 + c)^2 + c)^2 + c)^2 + c)^2 + c
Now you have the values for the first five iterations of v. Connect them with line segments to visualize the trajectory.
To find a complex number w that is in the prisoner set, we need to find a value that doesn't escape to infinity when iterated using the same f(z) function. You can select any value for w and apply the iterations in the same way as above. Choose a value that does not become infinitely large.
Finally, to find a complex number j in the Julia set, we need to find a value that lies on the boundary between the escape set and the prisoner set. This can be a bit trickier and usually requires more advanced mathematical techniques or computational algorithms. Depending on the complexity of the function f(z), finding a precise value for j may be challenging. However, if you want to approximate it, you can iterate the function for different initial values and observe where the trajectory starts exhibiting chaotic behavior.
I hope this explanation helps! Let me know if you have any other questions.
To solve this problem, let's break it down into steps:
Step 1: Choosing a complex number c.
You mentioned that you have already chosen c = 2i. That's a good choice to start with.
Step 2: Calculating the escape set.
For the function f(z) = z^2 + c, the escape set consists of all points that escape to infinity under iteration. To find a complex number v that is in the escape set, you can start with any complex number and iterate it using the function f(z) until it escapes to infinity. Let's choose v = 1 + i as an example.
Iteration 1: f(v) = (1 + i)^2 + 2i = 1 + 2i + i^2 + 2i = -1 + 4i
Iteration 2: f(f(v)) = (-1 + 4i)^2 + 2i = 1 + 8i - 16 + 8i + 2i = -15 + 18i
Iteration 3: f(f(f(v))) = (-15 + 18i)^2 + 2i = 225 - 324 + 540i = -99 + 540i
Iteration 4: f(f(f(f(v)))) = (-99 + 540i)^2 + 2i = 87583 + 107898i
As you can see, the magnitude of f(f(f(f(v)))) keeps growing without bound as the iterations go on. Therefore, v = 1 + i is in the escape set.
Step 3: Calculating the prisoner set.
The prisoner set consists of complex numbers that do not escape under iteration. To find a complex number w that is in the prisoner set, you can choose any complex number and iterate it using the function f(z) until it either converges or repeats a value. Let's choose w = 0 as an example.
Iteration 1: f(w) = 0^2 + 2i = 2i
Iteration 2: f(f(w)) = (2i)^2 + 2i = -4 + 2i + 2i = -4 + 4i
Iteration 3: f(f(f(w))) = (-4 + 4i)^2 + 2i = -12 - 16i
After three iterations, we can see that the values are not escaping to infinity, nor converging to a single value. Therefore, w = 0 is in the prisoner set.
Step 4: Graphing the first five iterations of v and w.
To graph the iterations, we will plot the complex numbers along the complex plane. Each iteration will be connected with line segments.
For v = 1 + i:
Iteration 1: (1 + i) -> (-1 + 4i)
Iteration 2: (-1 + 4i) -> (-15 + 18i)
Iteration 3: (-15 + 18i) -> (-99 + 540i)
Iteration 4: (-99 + 540i) -> (87583 + 107898i)
Iteration 5: (87583 + 107898i)
For w = 0:
Iteration 1: 0 -> 2i
Iteration 2: 2i -> -4 + 4i
Iteration 3: -4 + 4i -> -12 - 16i
Iteration 4: -12 - 16i -> 192 - 42i
Iteration 5: 192 - 42i
Now, you can plot the complex numbers on the complex plane and connect them with line segments to visualize the iterations.