# Periodic points of complex quadratic mappings

This article describes periodic points of some complex quadratic maps. A map is a formula for computing a value of a variable based on its own previous value or values; a quadratic map is one that involves the previous value raised to the powers one and two; and a complex map is one in which the variable is a complex number. A periodic point of a map is a value of the variable that occurs repeatedly after intervals of a fixed length.

This theory is applied in relation with the theories of Fatou and Julia sets.

## Definitions

Let

${\displaystyle f_{c}(z)=z^{2}+c\,}$

where ${\displaystyle z}$ and ${\displaystyle c}$ are complex-valued. (This ${\displaystyle \ f}$ is the complex quadratic mapping mentioned in the title.) This article explores the periodic points of this mapping - that is, the points that form a periodic cycle when ${\displaystyle \ f}$ is repeatedly applied to them.

Periodic points of a complex quadratic mapping of period ${\displaystyle \ p}$ are points ${\displaystyle \ z}$ of the dynamical plane such that :

where ${\displaystyle \ p}$ is the smallest positive integer.

We can introduce a new function:

so periodic points are zeros of function ${\displaystyle \ F_{p}(z,f)}$ :

which is a polynomial of degree ${\displaystyle \ =2^{p}}$

## Stability of periodic points (orbit) - multiplier

Stability index of periodic points along horizontal axis
boundaries of regions of parameter plane with attracting orbit of periods 1-6
Critical orbit of discrete dynamical system based on complex quadratic polynomial. It tends to weakly attracting fixed point with abs(multiplier)=0.99993612384259

The multiplier ( or eigenvalue, derivative ) ${\displaystyle m(f,z_{0})=\lambda \,}$ of rational map ${\displaystyle f\,}$ at fixed point ${\displaystyle z_{0}\,}$ is defined as :

Because the multiplier is the same at all periodic points, it can be called a multiplier of the periodic orbit.

The multiplier is:

A periodic point is :[2]

Where do periodic points belong?

• attracting is always in Fatou set
• repelling is in the Julia set
• Indifferent fixed points may be in one or the other.[3] Parabolic periodic point is in Julia set.

## Period-1 points (fixed points)

### Finite fixed points

Let us begin by finding all finite points left unchanged by 1 application of ${\displaystyle f}$. These are the points that satisfy ${\displaystyle \ f_{c}(z)=z}$. That is, we wish to solve

${\displaystyle z^{2}+c=z\,}$

which can be rewritten

${\displaystyle \ z^{2}-z+c=0.}$

Since this is an ordinary quadratic equation in 1 unknown, we can apply the standard quadratic solution formula. Look in any standard mathematics textbook, and you will find that there are two solutions of ${\displaystyle \ Ax^{2}+Bx+C=0}$ are given by

${\displaystyle x={\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}}$

In our case, we have ${\displaystyle A=1,B=-1,C=c}$, so we will write

${\displaystyle \alpha _{1}={\frac {1-{\sqrt {1-4c}}}{2}}}$ and ${\displaystyle \alpha _{2}={\frac {1+{\sqrt {1-4c}}}{2}}.}$

Since

${\displaystyle \alpha _{1}={\frac {1}{2}}-m}$ and ${\displaystyle \alpha _{2}={\frac {1}{2}}+m}$ where ${\displaystyle m={\frac {\sqrt {1-4c}}{2}}}$

It means that fixed points are symmetrical around ${\displaystyle z=1/2\,}$.

This image shows fixed points (both repelling)

#### Complex dynamics

Fixed points for c along horizontal axis
Fatou set for F(z)=z*z with marked fixed point

Here different notation is commonly used:[4]

${\displaystyle \alpha _{c}={\frac {1-{\sqrt {1-4c}}}{2}}}$ with multiplier ${\displaystyle \lambda _{\alpha _{c}}=1-{\sqrt {1-4c}}\,}$

and

${\displaystyle \beta _{c}={\frac {1+{\sqrt {1-4c}}}{2}}}$ with multiplier ${\displaystyle \lambda _{\beta _{c}}=1+{\sqrt {1-4c}}\,}$

Using Viète's formulas one can show that:

${\displaystyle \alpha _{c}+\beta _{c}=-{\frac {B}{A}}=1}$

Since derivative with respect to z is :

${\displaystyle P_{c}'(z)={\frac {d}{dz}}P_{c}(z)=2z}$

then

${\displaystyle P_{c}'(\alpha _{c})+P_{c}'(\beta _{c})=2\alpha _{c}+2\beta _{c}=2(\alpha _{c}+\beta _{c})=2\,}$

It implies that ${\displaystyle P_{c}\,}$ can have at most one attractive fixed point.

This points are distinguished by the facts that:

#### Special cases

An important case of the quadratic mapping is ${\displaystyle c=0}$. In this case, we get ${\displaystyle \alpha _{1}=0}$ and ${\displaystyle \alpha _{2}=1}$. In this case, 0 is a superattractive fixed point, and 1 belongs to the Julia set.

#### Only one fixed point

We might wonder what value ${\displaystyle c}$ should have to cause ${\displaystyle \alpha _{1}=\alpha _{2}}$. The answer is that this will happen exactly when ${\displaystyle 1-4c=0}$. This equation has 1 solution: ${\displaystyle c=1/4}$ (in which case, ${\displaystyle \alpha _{1}=\alpha _{2}=1/2}$). This is interesting, since ${\displaystyle c=1/4}$ is the largest positive, purely real value for which a finite attractor exists.

### Infinite fixed point

Then infinity is :

## Period-2 cycles

Bifurcation from period 1 to 2 for complex quadratic map

Suppose next that we wish to look at period-2 cycles. That is, we want to find two points ${\displaystyle \beta _{1}}$ and ${\displaystyle \beta _{2}}$ such that ${\displaystyle f_{c}(\beta _{1})=\beta _{2}}$, and ${\displaystyle f_{c}(\beta _{2})=\beta _{1}}$.

Let us start by writing ${\displaystyle f_{c}(f_{c}(\beta _{n}))=\beta _{n}}$, and see where trying to solve this leads.

${\displaystyle f_{c}(f_{c}(z))=(z^{2}+c)^{2}+c=z^{4}+2z^{2}c+c^{2}+c.\,}$

Thus, the equation we wish to solve is actually ${\displaystyle z^{4}+2cz^{2}-z+c^{2}+c=0}$.

This equation is a polynomial of degree 4, and so has 4 (possibly non-distinct) solutions. However, actually, we already know 2 of the solutions. They are ${\displaystyle \alpha _{1}}$ and ${\displaystyle \alpha _{2}}$, computed above. It is simple to see why this is; if these points are left unchanged by 1 application of ${\displaystyle f}$, then clearly they will be unchanged by 2 applications (or more).

Our 4th-order polynomial can therefore be factored in 2 ways :

### First method

${\displaystyle (z-\alpha _{1})(z-\alpha _{2})(z-\beta _{1})(z-\beta _{2})=0.\,}$

This expands directly as ${\displaystyle x^{4}-Ax^{3}+Bx^{2}-Cx+D=0}$ (note the alternating signs), where

${\displaystyle D=\alpha _{1}\alpha _{2}\beta _{1}\beta _{2}\,}$
${\displaystyle C=\alpha _{1}\alpha _{2}\beta _{1}+\alpha _{1}\alpha _{2}\beta _{2}+\alpha _{1}\beta _{1}\beta _{2}+\alpha _{2}\beta _{1}\beta _{2}\,}$
${\displaystyle B=\alpha _{1}\alpha _{2}+\alpha _{1}\beta _{1}+\alpha _{1}\beta _{2}+\alpha _{2}\beta _{1}+\alpha _{2}\beta _{2}+\beta _{1}\beta _{2}\,}$
${\displaystyle A=\alpha _{1}+\alpha _{2}+\beta _{1}+\beta _{2}.\,}$

We already have 2 solutions, and only need the other 2. This is as difficult as solving a quadratic polynomial. In particular, note that

${\displaystyle \alpha _{1}+\alpha _{2}={\frac {1-{\sqrt {1-4c}}}{2}}+{\frac {1+{\sqrt {1-4c}}}{2}}={\frac {1+1}{2}}=1}$

and

${\displaystyle \alpha _{1}\alpha _{2}={\frac {(1-{\sqrt {1-4c}})(1+{\sqrt {1-4c}})}{4}}={\frac {1^{2}-({\sqrt {1-4c}})^{2}}{4}}={\frac {1-1+4c}{4}}={\frac {4c}{4}}=c.}$

Adding these to the above, we get ${\displaystyle D=c\beta _{1}\beta _{2}}$ and ${\displaystyle A=1+\beta _{1}+\beta _{2}}$. Matching these against the coefficients from expanding ${\displaystyle f}$, we get

${\displaystyle D=c\beta _{1}\beta _{2}=c^{2}+c}$ and ${\displaystyle A=1+\beta _{1}+\beta _{2}=0.}$

From here, we construct a quadratic equation with ${\displaystyle A'=1,B=1,C=c+1}$ and apply the standard solution formula to get

${\displaystyle \beta _{1}={\frac {-1-{\sqrt {-3-4c}}}{2}}}$ and ${\displaystyle \beta _{2}={\frac {-1+{\sqrt {-3-4c}}}{2}}.}$

Closer examination shows (the formulas are a tad messy) that :

meaning these two points are the two halves of a single period-2 cycle.

### Second method of factorization

The roots of the first factor are the two fixed points ${\displaystyle z_{1,2}\,}$ . They are repelling outside the main cardioid.

The second factor has two roots

These two roots form period-2 orbit.[7]

#### Special cases

Again, let us look at ${\displaystyle c=0}$. Then

${\displaystyle \beta _{1}={\frac {-1-i{\sqrt {3}}}{2}}}$ and ${\displaystyle \beta _{2}={\frac {-1+i{\sqrt {3}}}{2}}}$

both of which are complex numbers. By doing a little algebra, we find ${\displaystyle |\beta _{1}|=|\beta _{2}|=1}$. Thus, both these points are "hiding" in the Julia set. Another special case is ${\displaystyle c=-1}$, which gives ${\displaystyle \beta _{1}=0}$ and ${\displaystyle \beta _{2}=-1}$. This gives the well-known superattractive cycle found in the largest period-2 lobe of the quadratic Mandelbrot set.

## Cycles for period>2

There is no general solution in radicals to polynomial equations of degree five or higher, so it must be computed using numerical methods.

## References

1. Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN 0-387-95151-2, p. 41
2. Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN 0-387-95151-2, page 99
3. Some Julia sets by Michael Becker
4. On the regular leaf space of the cauliflower by Tomoki Kawahira Source: Kodai Math. J. Volume 26, Number 2 (2003), 167-178.
5. Periodic attractor by Evgeny Demidov
6. R L Devaney, L Keen (Editor): Chaos and Fractals: The Mathematics Behind the Computer Graphics. Publisher: Amer Mathematical Society July 1989, ISBN 0-8218-0137-6 , ISBN 978-0-8218-0137-6
7. Period 2 orbit by Evgeny Demidov