source/reference:
https://www.youtube.com/channel/UCaTLkDn9_1Wy5TRYfVULYUw/playlists

Iteration of Complex Function

Iteration of Function

fⁿ is called the nth iterate of f

Let f(z)=z+1
f²(z)=f(f(z))=f(z+1)=(z+1)+1=z+2
f³(z)=f(f²(z))=f(z+2)=(z+2)+1=z+3
…
fⁿ(z)=f(fⁿ⁻¹(z))=f(z+(n-1))=(z+(n-1))+1=z+n

Let f(z)=3z
f²(z)=f(f(z))=f(3z)=3(3z)=3²z
f³(z)=f(f²(z))=f(3²z)=3(3²z)=3³z
…
fⁿ(z)=f(fⁿ⁻¹(z))=f(3ⁿ⁻¹z)=3(3ⁿ⁻¹z)=3ⁿz

Iteration of Quadratic Polynomials

The general quadratic polynomial is of the form p(z)=az²+bz+d, where constants a, b, d∊ℂ. However, for each triple of constants (a,b,d), there is exactly one constant c such that polynomial f(z)=z²+c behave the same as p(z)=az²+bz+d under iteration.

Given p(z)=az²+bz+d, where constants a, b, and d∊ℂ.
Define functons φ(z)=az+b
2 and f(z)=z²+c where c=ad+b
2-(b
2)²
 . 
⇒f(φ(z))=(φ(z))²+c=(az+b
2)²+ad+b
2-(b
2)²
 =a²z²+abz+ad+b
2=a(az²+bz+d)+b
2
⇒f(φ(z))=a(p(z))+b
2=φ(p(z))
⇒φ⁻¹(f(φ(z)))=p(z)
∴p(z)=φ⁻¹(f(φ(z))) for all z
rewrite as p=φ⁻¹∘f∘φ where ∘ means composed with
⇒p∘p=(φ⁻¹∘f∘φ)∘(φ⁻¹∘f∘φ)=φ⁻¹∘f∘φ∘φ⁻¹∘f∘φ=φ⁻¹∘f∘f∘φ
⇒p²=φ⁻¹∘f²∘φ
⇒p³=φ⁻¹∘f³∘φ 
…
⇒pⁿ=φ⁻¹∘fⁿ∘φ

It thus suffices to study the iteration of quadratic polynomials of the form f(z)=z²+c.

The Set of Iteration

The Julia set, named after the French mathematician Gaston Julia (1893-1978), of f(z)=z²+c is the set of all z∊ℂ for which the behavior of the iterates is "chaotic" in a neighborhood. the Fatou set, named after the French mathematician {oerre Fatpi (1878-1929), is the set of all z∊ℂ for which the iterates behave "normally" in a neighborhood.

The iterates of f behave normally near z if nearby point remain nearby under iteration. The iterates of behave chaotically at z if in any small neighborhood of z the behavior of the iterates depends sensitively on the initial point.

For example,

f(z)=z²+c. Let c=0, f(z)=z², then fⁿ(z)=z2ⁿ
Let z=re𝑖θ, then fⁿ(z)=r2ⁿe𝑖2ⁿθ
If |z|<1, then |fⁿ(z)|=|z|2ⁿ→0 as n→∞, so fⁿ(z)→0 as n→∞.
If |z|>1, then |fⁿ(z)|=|z|2ⁿ→∞ as n→∞, so fⁿ(z)→∞ as n→∞.
If |z|=1, then z=re𝑖θ=e𝑖θ, so fⁿ(z)=e𝑖2ⁿθ, thus |fⁿ(z)|=1 for all n
In any little disk around a point z with |z|=1, 
there are points w with |w|>1, for which fⁿ(w)→∞, 
and other points w with |w|<1, for which fⁿ(w)→0.
The unit circle {z:|z|=1} is thus 
the locus of chaotic behavior, whereas {z:|z|>1} with iterates attracted to ∞ 
and {z:|z|<1} with iterates attracted to 0 form the locus of normal behavior.

Julia set J(f)={z:|z|=1}, the unit circle. and Fatou set F(f)={z:|z|>1}∪{z:|z|<1}

The Basin of Attraction to ∞

Let f(z)=z²+c. Let A(∞)={z:fⁿ(z)→∞} "basin of attraction to ∞"

By theorem. The set A(∞) is open, connected and unbounded. It is contained in the Fatou set of f. The Julia set of f coincides with the boundary of A(∞) , which is a closed and bounded subset of ℂ.

Where

The Julia set J(f) is a closed and bounded set.
The Fatou set F(f) is open and unbounded and contains A(∞).
J(f)∩F(f)

Involution or conjugation

Let f(z)=z²-2. f²(z)=f(z²-2)=(z²-2)²-2. But it is hard to calculate and understand the iterates, fⁿ(z).

Conjugate f with

φ(w)=w+1
w,φ:{w:|w|>1}→ℂ\[-2,2].

f maps [-2,2] to[-2,2] and ℂ\[-2,2] to ℂ\[-2,2]. Thus look at φ⁻¹∘f∘φ.

Let f(z)=z²-2. φ(w)=w+1
w,
⇒f(φ(w))=(φ(w))²-2=(w+1
w)²-2=w²+2w
w+1
w²-2=w²+1
w²=φ(w²)
⇒φ⁻¹(f(φ(w)))=w²
Let g(w)=w²
⇒φ⁻¹(f(φ(w)))=g(w)
⇒ f(φ(w))=φ(g(w))
Let φ(w)=z⇒w=φ⁻¹(z)
⇒ f(z)=φ(g(φ⁻¹(z)))

Thus, on ℂ\[-2,2], the function f(z)=z²-2 behaves like g(w)=w² behaves on the exterior of the closed unit disk. Since the iterates gⁿ(w) tend to ∞ for |w|>1. therefore fⁿ(z)→∞ as n→∞ for all z∊ℂ\[-2,2]. Thus A(∞)=ℂ\[-2,2], and thus J(f)=[-2,2], the closed interval from -2 to 2 on the real axis. Both the Julia sets of f(z)=z² and f(z)=z²-2 are the only smooth Julia set examples amongst all f(z)=z²+c.

Julia Sets

For f(z)=z²+c, where c∊ℂ,the Julia set of f is the boundary of A(∞), where A(∞) is the "basin of attraction to infinity", i.e. A(∞)={z∊ℂ:fⁿ(z)→∞ as n→∞}. Let K(f) be the filled-in Julia set of f for which z∊ℂ and {fⁿ(z)} stays bounded, i.e. K(f)={z∊ℂ : {fⁿ(z)} is bounded}. For examples

f(z)=z² (c=0): Then K(f)={z:|z|≤1} and J(f)={z:|z|=1}
f(z)=z²-2 (c=-2): Then K(f)=[-2,2] and J(f)=[-2,2]

For f(z)=z²-1 (c=-1)

Let f(z)=z²-1
z=0: f(0)=-1, f²(0)=f(-1)=0, f³(0)=f(0)=-1, f⁴(0)=0, …. The orbit is thus 0, -1, 0, -1, 0, -1, …. 
This is called a periodic orbit, and 0 is a periodic point of period 2. clearly, 0∊K(f).
z=1: 1, 0, -1, 0, -1, 0, -1, …. So 1 isnot itself a periodic point (the orbit never returen to 1), 
but it's the next best thing;it runs into a periodic orbit. It thus is called 1 a pre-periodic point. 
Again, 1∊K(f).
z=1+√5
2:(1+√5
2)²
 -1=1+2√5+5
4-1=2+2√5
4=1+√5
2=z.
The point 1+√5
2is a fixed point of f and thus belongs to K(f) as well.

Other z

z=-2: -2, 3, 8, 63,  …, →∞. Thus -2∊A(∞).
z=: -2, 3, 8, 63,  …, →∞. Thus -2∊A(∞).

First, if z lies on the real axis and is smaller than -(1+√5
2) or larger than (1+√5
2), then z∊A(∞).

More generally, if z∊ℂ and |z|>1+√5
2, then z∊A(∞).

And more generally, if z∊ℂ and |fⁿ(z)|>1+√5
2 for some n, then z∊A(∞).

A similar condition holds for general quadratic polynomials f(z)=z²+c,

By theorem, Let f(z)=z²+c, and let R= 1+√(1+4|c|)
2. Let z₀∊ℂ. If for some n>0, there are |fⁿ(z)|→∞as n →∞, i.e. z₀∊A(∞), so z₀∉K(f).

Computer Algorithm finding the Filled-In Julia Sets

Computer algorithm:

Given c, let R= 1+√(1+4|c|)
2.
Choose a window W={x+𝑖y: a≤x≤b, c≤y≤d} to display. If you want to see the entire Julia set, you'd want BR(0)⊂W. But at some point it also may be interesting to zoom into the Julia set!
The computer can't keep checking all iterates fⁿ(z), at some point it'll have to stop. So pick a largest number, maxiter, up to which it will check. The larger this number, the more accurate your picture will, get, but the slower the calculation will be.
For each pixel on your screen, choose a point z in your window W, corresponding to that pixel.
Calculate the iterates of z: f(z), f(f(z)), f(f(f(z))), …. If one of these iterates satisfies that |fⁿ(z)|>R, color the initial pixel white.
If you reach the maximum number of iterations, maxiter, without having left BR(0), there's a good chance that z belongs to K(f). Color this pixel black.

Prettier Pictures

The pictures of the filled-in Julia sets produced so far are only black and white. Fro colorful pictures, choose colors c₀, c₁, c₂, …, cmaxiter for A(∞) as well as a color for the filled-in Julia set. If |z|>R, color the corresponding pixel with color c₀. Otherwise, if |f(z)|>R, color the corresponding pixel with color c₁. Otherwise, if |f(f(z))|>R, color the corresponding pixel with color c₂. …. Otherwise, i.e. if |fⁿ(z)|>R for all n≤maxiter, color the corresponding pixel with corresponding color cₙ for the filled-in Julia set. The larger constant maxiter, the more detail of the pictures, but the longer the calculation will take. For small windows (i.e. zooms into the Julia set), a higher precision is required to avoid round-off errors.

The Mandelbrot Set

The Julia set of f(z)=z²+c is either "in one piece" or "totally dusty"

By definition. The Mandelbrot set M is the set of all parameters c∊ℂ for which the Julia set J(f) of f(z)=z²+c is connected in one piece. That is M={c∊ℂ: J(z²+c) is connected}

Note: The Mandelbrot set is a subset of the parameter space (the space of all possible c-values), whereas Julia sets are sets of z-values.

Examples

0∊M since J(z²)={z"|z|=1}
-2∊M since J(z²-2)=[-2,2]
¼∊M according to picture of J(z²+¼)
1∉M according to picture of J(z²+1)

By theorem. Let f(z)=z²+c. Then J(f) is connected if and only if z=0 does not belong to A(∞), that is if and only if the orbit {fⁿ(0)} remains bounded under iteration.

In fact, it is possible to show the following:

By theorem. A complex number c belongs to M if and only if |fⁿ(0)|≤2 for all n≥1 (where f(z)=z²+c).

Computer Algorithm to Plot the Mandelbrot Set

Computer algorithm

Choose a window W={cx+𝑖cy:cxmin≤cx≤cxmax , cymin≤cy≤cymax } to display. If you want to see the entire Mandelbrot set, you'd want something like W={c: -2≤cx≤0.75 , -1.5≤cy≤1.5 }
As before, pick a largest number of iterations, maxiter. The larger this number, the more accurate the picture will get, but the slower the calculation will be.
For each pixel on the screen, choose a parameter c in the window W, corresponding to that pixel. We'll look at the corresponding polynomial f(z)=z²+c.
Calculate the iterates of 0 under this polynomial: f(0)=c, f(f0))=c²+c, f(f(f(0)))=(c²+c)²+c, …. If one of these iterates satisfies that |fⁿ(0)|>2, color the initial pixel white.
If you reach the maximum number of iterations, maxiter, without having left B₂(0), there's a good chance that the parameter c belongs to the Mandelbrot set M. Color this pixel black.

A Prettier Picture

Again, you can use different colors for those parameters c∊ℂ for which 0 escapes to infinity under iteration, depending on how quickly the escape happens"

I |f(0)|=|c|>2, color the corresponding pixel in color zero.
Otherwise, if |f(f0))|=|c²+c|>2, color the corresponding pixel in color one.
Otherwise, if |f(f(f0)))|=|(c²+c)²+c|>2, color the corresponding pixel in color two.
….
If |fⁿ(0)|≤2 for all n≤maxiter, color the pixel corresponding to c black (or whatever other color you choose for your M).

Zooming into the Mandelbrot set and coloring parameters by escape time yields beautiful pictures.

Properties of the Mandelbrot Set

M is a connected set (Douadym Hubbard, 1982)
M is contained in the disk of radius 2, centered at zero.
The boundary of M is very intricate- this is where you'll find the most beatutiful zooms.
Moreover, for c-values near the boundary of M, their Julia sets have many different patterns. Here are some examples:
The boundary of the main cardioid is given by c=½e𝑖θ-¼e2𝑖θ,0≤θ<2π.
Writing θ=2πα, 0≤α<1. we can distinguish whether α is a rational or an irrational number.
Rational α:
- Let c=½e2π𝑖α-¼e4π𝑖α where ,0≤α<1 is a rational number. Then α is of the form p/q.
- The parameter c is an attachment point of another "bud" to the Mandelbrot set, and the Julia set for parameter values within the bud.
- Example: α=½. Then c=½eπ𝑖-¼e2π𝑖=-½-¼=-0.75. A picture for J(z²-0.75)
Irrational α:
- Let c=½e2π𝑖α-¼e4π𝑖α where ,0≤α<1 is a irrational number. Thus there are no values p and q such that α is of the form p/q.
- Julia sets for such values look more intricate and come in several "flavors"!
- For example: α= 1+√5
  2. Then c≡-0.390540870218401…-0.586787907346969…i. For f(z)=z²+c, the interior of K(f) has a so-called "Siegel disk", in which iteration looks like a rotation by angle α.

Misiurewicz Points

Many points in the boundary of M are co-called Misiurewicz points:

A point c∊ℂ is called a Misiurewicz point if the orbit of 0 under f(z)=z²+c is pre-periodic, but not periodic.
Example: c=𝑖: Then f(z)=z²+𝑖 and the orbit of 0 under f is 0, 𝑖, -1+𝑖, -𝑖, -1+𝑖, -𝑖, ….
Clearly, Misiurewicz points c belong to M since the orbit of 0 under f(z)=z²+c is bounded.
Properties:
- Let c be a Misiurewicz point. Then J(f)=K(f), i.e. K(f) has no interior.
- Misiurewicz points are dense in ∂M.
- The Mandelbrot set is self-similar under magnification near Misiurewicz points. Note: The Mandelbrot is "quasi-self-similar" everywhere: small, slightly different versions of itself can be found at arbitrary small scales.

Big Open Conjecture

Here is one of the big outstanding conjectures in the field of complex dynamics:

By Conjecture. The Mandelbrot set is locally connected, that is, for every c∊M and every open set V with c∊V, there exists an open set U such that c∊U⊂V and U∩M is connected.