Classify some fixed points

test-prep

#1

Find and classify the fixed points of each of the following functions:

  • \displaystyle f(z) = \frac{z^2}{2}+\frac{5 z}{2}+1
  • \displaystyle f(z) = z^2+(2+i) z+i
  • \displaystyle f(z) = \frac{1}{3}z + 6z^3

#2
  1. z = \frac{z^2}{2}+\frac{5 z}{2}+1, so 0= \frac{z^2}{2}+\frac{3 z}{2}+1. By applying the quadratic formula, we find the roots are given by \frac{-\frac{3}{2}\pm\sqrt{\left(\frac{3}{2}\right)^2 -4\left(\frac{1}{2}\right)}}{2\left(\frac{1}{2} \right)}=-\frac{3}{2}\pm\frac{1}{2}. Thus, our fixed points are z=-1,-2.

  2. z = z^2+(2+i)z+i, so 0= z^2+(1+i)z+i. By applying the quadratic formula, we find the roots are given by \frac{-(1+i)\pm\sqrt{(1+i)^2 -4i}}{2}. This simplifies to \frac{-(1+i)\pm(1-i)}{2}, since \sqrt{-2i}=\sqrt{1-2i-1} = \sqrt{1-2i+i^2}=\sqrt{(1-i)^2}=(1-i). Thus, our fixed points are z=-1,-i.

  3. z = 6z^3+\frac{z}{3}, so we can see that clearly zero is a solution. Continuing, 0= 6z^2+\frac{1}{3}, so when we apply the quadratic formula we get that the roots are \frac{\pm\sqrt{-4(6)(\frac{1}{3})}}{2(6)}, which simplifies to \frac{\pm i\sqrt{2}}{6}. Thus our fixed points are z=0, \frac{i\sqrt{2}}{6}, \frac{-i\sqrt{2}}{6}.


#3

@MikhailTal
I think these answers are partially good but not complete. When we are asked to classify fixed points in complex dynamics, we mean classify them them as (super-) attractive, repulsive, or neutral. For example, the first function

f(z) = \frac{1}{2}z^2 + \frac{5}{2}z+1

certainly fixes -1 and -2. In addition,

f'(z) = z + \frac{5}{2}

so that f(-1) = 3/2 and f(-2)=1/2. Thus, -1 is an attractive fixed point while -2 is a repulsive fixed point.