# 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

$f_{c}(z)=z^{2}+c\,$ We can introduce a new function:

## Stability of periodic points (orbit) - multiplier

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 :

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. 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 $f$ . These are the points that satisfy $\ f_{c}(z)=z$ . That is, we wish to solve

$z^{2}+c=z\,$ which can be rewritten

$\ z^{2}-z+c=0.$ 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 $\ Ax^{2}+Bx+C=0$ are given by

$x={\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}$ $\alpha _{1}={\frac {1-{\sqrt {1-4c}}}{2}}$ and $\alpha _{2}={\frac {1+{\sqrt {1-4c}}}{2}}.$ Since

$\alpha _{1}={\frac {1}{2}}-m$ and $\alpha _{2}={\frac {1}{2}}+m$ where $m={\frac {\sqrt {1-4c}}{2}}$ It means that fixed points are symmetrical around $z=1/2\,$ .

#### Complex dynamics

Here different notation is commonly used:

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

$\beta _{c}={\frac {1+{\sqrt {1-4c}}}{2}}$ with multiplier $\lambda _{\beta _{c}}=1+{\sqrt {1-4c}}\,$ Using Viète's formulas one can show that:

$\alpha _{c}+\beta _{c}=-{\frac {B}{A}}=1$ Since derivative with respect to z is :

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

$P_{c}'(\alpha _{c})+P_{c}'(\beta _{c})=2\alpha _{c}+2\beta _{c}=2(\alpha _{c}+\beta _{c})=2\,$ It implies that $P_{c}\,$ can have at most one attractive fixed point.

This points are distinguished by the facts that:

### Infinite fixed point

Then infinity is :

## Period-2 cycles

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

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

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 $\alpha _{1}$ and $\alpha _{2}$ , computed above. It is simple to see why this is; if these points are left unchanged by 1 application of $f$ , then clearly they will be unchanged by 2 applications (or more).

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

### First method

$(z-\alpha _{1})(z-\alpha _{2})(z-\beta _{1})(z-\beta _{2})=0.\,$ This expands directly as $x^{4}-Ax^{3}+Bx^{2}-Cx+D=0$ (note the alternating signs), where

$D=\alpha _{1}\alpha _{2}\beta _{1}\beta _{2}\,$ $C=\alpha _{1}\alpha _{2}\beta _{1}+\alpha _{1}\alpha _{2}\beta _{2}+\alpha _{1}\beta _{1}\beta _{2}+\alpha _{2}\beta _{1}\beta _{2}\,$ $B=\alpha _{1}\alpha _{2}+\alpha _{1}\beta _{1}+\alpha _{1}\beta _{2}+\alpha _{2}\beta _{1}+\alpha _{2}\beta _{2}+\beta _{1}\beta _{2}\,$ $A=\alpha _{1}+\alpha _{2}+\beta _{1}+\beta _{2}.\,$ We already have 2 solutions, and only need the other 2. This is as difficult as solving a quadratic polynomial. In particular, note that

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

$\alpha _{1}\alpha _{2}={\frac {(1-{\sqrt {1-4c}})(1+{\sqrt {1-4c}})}{4}}={\frac {1^{2}-({\sqrt {1-4c}})^{2}}{4}}={\frac {1-1+4c}{4}}={\frac {4c}{4}}=c.$ $D=c\beta _{1}\beta _{2}=c^{2}+c$ and $A=1+\beta _{1}+\beta _{2}=0.$ From here, we construct a quadratic equation with $A'=1,B=1,C=c+1$ and apply the standard solution formula to get

$\beta _{1}={\frac {-1-{\sqrt {-3-4c}}}{2}}$ and $\beta _{2}={\frac {-1+{\sqrt {-3-4c}}}{2}}.$ 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 $z_{1,2}\,$ . They are repelling outside the main cardioid.

The second factor has two roots

These two roots form period-2 orbit.

#### Special cases

$\beta _{1}={\frac {-1-i{\sqrt {3}}}{2}}$ and $\beta _{2}={\frac {-1+i{\sqrt {3}}}{2}}$ both of which are complex numbers. By doing a little algebra, we find $|\beta _{1}|=|\beta _{2}|=1$ . Thus, both these points are "hiding" in the Julia set. Another special case is $c=-1$ , which gives $\beta _{1}=0$ and $\beta _{2}=-1$ . 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.