Let f(z) = i z^2 + z + 2i.
- Find and classify the fixed points of f.
- Write down an equation that the points of period 2 must satisfy.
Let f(z) = i z^2 + z + 2i.
Let f(z)=iz^2 + z + 2i
z_{0} is a fixed point of f(z) if f(z_{0})=z_{0}.
f(z_{0})=z_{0}= iz_{0}^2 + z_{0} +2i
iz_{0}^2 +2i=0
z_{0}^2=-2, so the fixed points of f are z_{0} = -\sqrt{2}i and z_{0} =\sqrt{2}i.
f'(z)=2iz+1
|f'(-\sqrt{2}i)|=|2\sqrt{2}+1|>1, so the fixed point at -\sqrt{2}i is repulsive.
|f'(\sqrt{2}i)|=|-2\sqrt{2}+1|>1, so the fixed point at \sqrt{2}i is repulsive.
We can say a point z_{0} is a periodic with period n under iteration of f if f(z_{0})^n=z_{0} and f(z_{0})^k \neq z_{0} where k=1,2,...,n-1.
f(z_{0})^k \neq z_{0} can only be f(z_{0}) = z_{0}, which means that points periodic with period 2 can not be fixed points of f.
We can say F=f^n=f^2, and points periodic with period 2 for f will be fixed points of F. So the equation that points of period 2 must satisfy is F(z_{0})=z_{0} or F(z_{0})= (iz_{0}^2 + z_{0} + 2i)^2=z_{0}.