Periodic points of complex quadratic mappings

Let us begin by finding all finite points left unchanged by one 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 as

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

Since this is an ordinary quadratic equation in one unknown, we can apply the standard quadratic solution formula:

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

So for ${\displaystyle c\in {\mathbb {C} }\setminus \{1/4\}}$ we have two finite fixed points ${\displaystyle \alpha _{1}\,}$ and ${\displaystyle \alpha _{2}\,}$.

Since

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

then ${\displaystyle \alpha _{1}+\alpha _{2}=1}$.

Thus fixed points are symmetrical around ${\displaystyle z=1/2}$.

This image shows fixed points (both repelling)

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:

${\displaystyle \alpha _{c}+\beta _{c}=1.}$

Since the derivative with respect to z is

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

then

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

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

These points are distinguished by the facts that:

• ${\displaystyle \beta _{c}\,}$ is:
  • the landing point of the external ray for angle=0 for ${\displaystyle c\in M\setminus \left\{{\frac {1}{4}}\right\}}$
  • 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).[5]
• ${\displaystyle \alpha _{c}\,}$ is:
  • the landing point of several rays
  • attracting when ${\displaystyle c}$ 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 ${\displaystyle c}$

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.

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

We can extend the complex plane ${\displaystyle \mathbb {C} }$ to the Riemann sphere (extended complex plane) ${\displaystyle \mathbb {\hat {C}} }$ by adding infinity :

${\displaystyle \mathbb {\hat {C}} =\mathbb {C} \cup \{\infty \}}$

and extending polynomial ${\displaystyle f_{c}\,}$such that ${\displaystyle f_{c}(\infty )=\infty .\,}$

Then infinity is :

• superattracting
• a fixed point of polynomial ${\displaystyle f_{c}:\,}$[6]

${\displaystyle f_{c}(\infty )=\infty =f_{c}^{-1}(\infty ).\,}$

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

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 two solutions, and only need the other two. Hence the problem is equivalent to 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 this, we easily get

${\displaystyle \beta _{1}\beta _{2}=c+1}$ and ${\displaystyle \beta _{1}+\beta _{2}=-1}$.

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 that :

${\displaystyle f_{c}(\beta _{1})=\beta _{2}}$ and ${\displaystyle f_{c}(\beta _{2})=\beta _{1},}$

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

We can factor the quartic by using polynomial long division to divide out the factors ${\displaystyle (z-\alpha _{1})}$ and ${\displaystyle (z-\alpha _{2}),}$ which account for the two fixed points ${\displaystyle \alpha _{1}}$ and ${\displaystyle \alpha _{2}}$ (whose values were given earlier and which still remain at the fixed point after two iterations):

${\displaystyle (z^{2}+c)^{2}+c-z=(z^{2}+c-z)(z^{2}+z+c+1).\,}$

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

${\displaystyle {\frac {-1\pm {\sqrt {-3-4c}}}{2}}.\,}$

These two roots, which are the same as those found by the first method, form the period-2 orbit.[7]

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. We have ${\displaystyle |\beta _{1}|=|\beta _{2}|=1}$. 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.