TITLE: Recreational math problem: $f(x)=\frac{x^2}{2x+101}$
QUESTION [6 upvotes]: Here is a math problem (just for fun) for the residents of MSE to enjoy:

Let $f(z)$ be defined as 
  $$f(z)=\frac{z^2}{2z+101}$$
  Find a 4-cycle of $f$ - that is, find four distinct complex numbers $z_1,z_2,z_3,z_4$ such that
  $$f(z_1)=z_2$$
  $$f(z_2)=z_3$$
  $$f(z_3)=z_4$$
  $$f(z_4)=z_1$$

The answer (my answer, at least) is a bit messy, so watch out! After a correct answer is posted, I will post my solution method.
Enjoy!

REPLY [3 votes]: Here is my solution, as promised. First, I restate the problem:

Let $f(z)$ be defined as 
  $$f(z)=\frac{z^2}{2z+101}$$
  Find a 4-cycle of $f$ - that is, find four distinct complex numbers $z_1,z_2,z_3,z_4$ such that
  $$f(z_1)=z_2$$
  $$f(z_2)=z_3$$
  $$f(z_3)=z_4$$
  $$f(z_4)=z_1$$

Notice that $f$ is equal to
$$f(z)=\frac{1}{101\bigg(\frac{1}{z}+\frac{1}{101}\bigg)^{2}-\frac{1}{101}}$$
from which it easily follows that
$$f^{\circ n}(z)=\frac{1}{101\bigg(\frac{1}{z}+\frac{1}{101}\bigg)^{2^n}-\frac{1}{101}}$$
where $f^{\circ n}$ represents $f$ composed $n$ times. In order for $f$ to have a $4$-cycle, $f^{\circ n}(z_1)$ must be a periodic function of $n$ with period $4$. This occurs if and only if
$$\bigg(\frac{1}{z_1}+\frac{1}{101}\bigg)^{2^n}$$
is also a periodic function of $n$. Now recall some properties of the complex roots of unity. If $\omega_k$ is defined as
$$\omega_k=\cos\frac{2\pi}{k}+i\sin\frac{2\pi}{k}$$
then an elementary property is that
$$\omega_k^p=\omega_k^{p\bmod k}$$
Now notice that $2^n\bmod 5$ is periodic with period $4$, with
$$2^{4n}\bmod 5 =1$$
$$2^{4n+1}\bmod 5 =2$$
$$2^{4n+2}\bmod 5 =4$$
$$2^{4n+3}\bmod 5 =3$$
Thus, $f^{\circ n}(z_1)$ should be periodic if we let
$$\frac{1}{z_1}+\frac{1}{101}=\omega_5$$
so that
$$\frac{1}{z_2}+\frac{1}{101}=\omega_5^2$$
$$\frac{1}{z_3}+\frac{1}{101}=\omega_5^4$$
$$\frac{1}{z_4}+\frac{1}{101}=\omega_5^3$$
giving us the values
$$z_1=\frac{101}{101\omega_5-1}$$
$$z_2=\frac{101}{101\omega_5^2-1}$$
$$z_3=\frac{101}{101\omega_5^4-1}$$
$$z_4=\frac{101}{101\omega_5^3-1}$$
...which can be simplified, of course, but you guys have already done that part for me.