# 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 and the parameters are complex numbers. A periodic point of a map is a value of the variable that occurs repeatedly after intervals of a fixed length.
These periodic points play a role in the theories of Fatou and Julia sets.

## Definitions

Let
    $f_{c}(z)=z^{2}+c\,$
be the complex quadric mapping, where $z$ and $c$ are complex numbers.
Notationally, $f_{c}^{(k)}(z)$ is the $k$-fold composition of $f_{c}$ with itself (not to be confused with the $k$th derivative of $f_{c}$)—that is, the value after the k-th iteration of the function $f_{c}.$ Thus
    $f_{c}^{(k)}(z)=f_{c}(f_{c}^{(k-1)}(z)).$
Periodic points of a complex quadratic mapping of period $p$ are points $z$ of the dynamical plane such that
    $f_{c}^{(p)}(z)=z,$
where $p$ is the smallest positive integer for which the equation holds at that z.
We can introduce a new function:
    $F_{p}(z,f)=f_{c}^{(p)}(z)-z,$
so periodic points are zeros of function $F_{p}(z,f)$: points z satisfying
    $F_{p}(z,f)=0,$
which is a polynomial of degree $2^{p}.$

## Number of periodic points

The degree of the polynomial $F_{p}(z,f)$ describing periodic points is $d=2^{p}$ so it has exactly $d=2^{p}$ complex roots (= periodic points), counted with multiplicity.

## Stability of periodic points (orbit) - multiplier

The multiplier (or eigenvalue, derivative) $m(f^{p},z_{0})=\lambda$ of a rational map $f$ iterated $p$ times at cyclic point $z_{0}$ is defined as:
    $m(f^{p},z_{0})=\lambda ={\begin{cases}f^{p\prime }(z_{0}),&{\mbox{if }}z_{0}\neq \infty \\{\frac {1}{f^{p\prime }(z_{0})}},&{\mbox{if }}z_{0}=\infty \end{cases}}$
where $f^{p\prime }(z_{0})$ is the first derivative of $f^{p}$ with respect to $z$ at $z_{0}$.
Because the multiplier is the same at all periodic points on a given orbit, it is called a multiplier of the periodic orbit.
The multiplier is:
* a complex number;
* invariant under conjugation of any rational map at its fixed point;
* used to check stability of periodic (also fixed) points with stability index $abs(\lambda ).\,$
A periodic point is
* attracting when $abs(\lambda )<1;$
** super-attracting when $abs(\lambda )=0;$
** attracting but not super-attracting when $0<abs(\lambda )<1;$
* indifferent when $abs(\lambda )=1;$
** rationally indifferent or parabolic if $\lambda$ is a root of unity;
** irrationally indifferent if $abs(\lambda )=1$ but multiplier is not a root of unity;
* repelling when $abs(\lambda )>1.$
Periodic points
* that are attracting are always in the Fatou set;
* that are repelling are in the Julia set;
* that are indifferent fixed points may be in one or the other. A parabolic periodic point is in the Julia set.

## Period-1 points (fixed points)
### Finite fixed points

Let us begin by finding all finite points left unchanged by one 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 as
    $\ z^{2}-z+c=0.$
Since this is an ordinary quadratic equation in one unknown, we can apply the standard quadratic solution formula:
    $\alpha _{1}={\frac {1-{\sqrt {1-4c}}}{2}}$ and $\alpha _{2}={\frac {1+{\sqrt {1-4c}}}{2}}.$
So for $c\in \mathbb {C} \setminus \{1/4\}$ we have two finite fixed points $\alpha _{1}$ and $\alpha _{2}$.
Since
    $\alpha _{1}={\frac {1}{2}}-m$ and $\alpha _{2}={\frac {1}{2}}+m$ where $m={\frac {\sqrt {1-4c}}{2}},$
we have $\alpha _{1}+\alpha _{2}=1$.
Thus fixed points are symmetrical about $z=1/2$.

#### 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 $\lambda _{\beta _{c}}=1+{\sqrt {1-4c}}.$
Again we have
    $\alpha _{c}+\beta _{c}=1.$
Since the derivative with respect to z is
    $P_{c}'(z)={\frac {d}{dz}}P_{c}(z)=2z,$
we have
    $P_{c}'(\alpha _{c})+P_{c}'(\beta _{c})=2\alpha _{c}+2\beta _{c}=2(\alpha _{c}+\beta _{c})=2.$
This implies that $P_{c}$ can have at most one attractive fixed point.
These points are distinguished by the facts that:
* $\beta _{c}$ is:
** the landing point of the external ray for angle=0 for $c\in M\setminus \left\{1/4\right\}$
** the most repelling fixed point of the Julia set
** the one on the right (whenever fixed point are not symmetrical around the real axis), it is the extreme right point for connected Julia sets (except for cauliflower).
* $\alpha _{c}$ is:
** the landing point of several rays
** attracting when $c$ is in the main cardioid of the Mandelbrot set, in which case it is in the interior of a filled-in Julia set, and therefore belongs to the Fatou set (strictly to the basin of attraction of finite fixed point)
** parabolic at the root point of the limb of the Mandelbrot set
** repelling for other values of $c$

#### Special cases

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

#### Only one fixed point

We have $\alpha _{1}=\alpha _{2}$ exactly when $1-4c=0.$ This equation has one solution, $c=1/4,$ in which case $\alpha _{1}=\alpha _{2}=1/2$. In fact $c=1/4$ is the largest positive, purely real value for which a finite attractor exists.

### Infinite fixed point

We can extend the complex plane $\mathbb {C}$ to the Riemann sphere (extended complex plane) $\mathbb {\hat {C}}$ by adding infinity:
    $\mathbb {\hat {C}} =\mathbb {C} \cup \{\infty \}$
and extend $f_{c}$ such that $f_{c}(\infty )=\infty .$
Then infinity is:
* superattracting
* a fixed point of $f_{c}$:$f_{c}(\infty )=\infty =f_{c}^{-1}(\infty ).$

## Period-2 cycles

Period-2 cycles are two distinct points $\beta _{1}$ and $\beta _{2}$ such that $f_{c}(\beta _{1})=\beta _{2}$ and $f_{c}(\beta _{2})=\beta _{1}$, and hence
    $f_{c}(f_{c}(\beta _{n}))=\beta _{n}$
for $n\in \{1,2\}$:
    $f_{c}(f_{c}(z))=(z^{2}+c)^{2}+c=z^{4}+2cz^{2}+c^{2}+c.$
Equating this to z, we obtain
    $z^{4}+2cz^{2}-z+c^{2}+c=0.$
This equation is a polynomial of degree 4, and so has four (possibly non-distinct) solutions. However, we already know two of the solutions. They are $\alpha _{1}$ and $\alpha _{2}$, computed above, since if these points are left unchanged by one application of $f$, then clearly they will be unchanged by more than one application of $f$.
Our 4th-order polynomial can therefore be factored in 2 ways:

### First method of factorization

    $(z-\alpha _{1})(z-\alpha _{2})(z-\beta _{1})(z-\beta _{2})=0.\,$
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 two solutions, and only need the other two. Hence the problem is equivalent to 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
    $\alpha _{1}\alpha _{2}={\frac {(1-{\sqrt {1-4c}})(1+{\sqrt {1-4c}})}{4}}={\frac {1^{2}-({\sqrt {1-4c}})^{2}}{4}}={\frac {1-1+4c}{4}}={\frac {4c}{4}}=c.$
Adding these to the above, we get $D=c\beta _{1}\beta _{2}$ and $A=1+\beta _{1}+\beta _{2}$. Matching these against the coefficients from expanding $f$, we get
    $D=c\beta _{1}\beta _{2}=c^{2}+c$ and $A=1+\beta _{1}+\beta _{2}=0.$
From this, we easily get
    $\beta _{1}\beta _{2}=c+1$ and $\beta _{1}+\beta _{2}=-1$.
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 that:
    $f_{c}(\beta _{1})=\beta _{2}$ and $f_{c}(\beta _{2})=\beta _{1},$
meaning these two points are the two points on a single period-2 cycle.

### Second method of factorization

We can factor the quartic by using polynomial long division to divide out the factors $(z-\alpha _{1})$ and $(z-\alpha _{2}),$ which account for the two fixed points $\alpha _{1}$ and $\alpha _{2}$ (whose values were given earlier and which still remain at the fixed point after two iterations):
    $(z^{2}+c)^{2}+c-z=(z^{2}+c-z)(z^{2}+z+c+1).\,$
The roots of the first factor are the two fixed points. They are repelling outside the main cardioid.
The second factor has the two roots
    ${\frac {-1\pm {\sqrt {-3-4c}}}{2}}.\,$
These two roots, which are the same as those found by the first method, form the period-2 orbit.

#### Special cases

Again, let us look at $c=0$. Then
    $\beta _{1}={\frac {-1-i{\sqrt {3}}}{2}}$ and $\beta _{2}={\frac {-1+i{\sqrt {3}}}{2}},$
both of which are complex numbers. We have $|\beta _{1}|=|\beta _{2}|=1$. 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 greater than 2

The degree of the equation $f^{(n)}(z)=z$ is 2$^n$; thus for example, to find the points on a 3-cycle we would need to solve an equation of degree 8. After factoring out the factors giving the two fixed points, we would have a sixth degree equation.
There is no general solution in radicals to polynomial equations of degree five or higher, so the points on a cycle of period greater than 2 must in general be computed using numerical methods. However, in the specific case of period 4 the cyclical points have lengthy expressions in radicals.
In the case c = –2, trigonometric solutions exist for the periodic points of all periods. The case $z_{n+1}=z_{n}^{2}-2$ is equivalent to the logistic map case r = 4: $x_{n+1}=4x_{n}(1-x_{n}).$ Here the equivalence is given by $z=2-4x.$ One of the k-cycles of the logistic variable x (all of which cycles are repelling) is
    $\sin ^{2}\left({\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2\cdot {\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2^{2}\cdot {\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2^{3}\cdot {\frac {2\pi }{2^{k}-1}}\right),\dots ,\sin ^{2}\left(2^{k-1}{\frac {2\pi }{2^{k}-1}}\right).$