title: Periodic points of complex quadratic mappings
text: # 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
  
      $\sin ^{2}\left({\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2\cdot {\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2^{2}\cdot {\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2^{3}\cdot {\frac {2\pi }{2^{k}-1}}\right),\dots ,\sin ^{2}\left(2^{k-1}{\frac {2\pi }{2^{k}-1}}\right).$
  