Consider fix point of \( R(z) = z^2 – z \) .

Which is the solution of $$ R(z) = z \\ \Rightarrow z^2 – z =z \\ \Rightarrow z^2 – 2z =0 \\ \Rightarrow z(z -2) =0 $$

Now , consider the fix point of \( R^2(z) \) . \( \\ \) $$ i.e. R^2(z) = R . R(z) = R(z^2 -z)\\ \Rightarrow (z^2 -z)^2 – z^2 +z =z \\ \Rightarrow z^4 -2z^3 = 0 \\ \Rightarrow z^3(z- 2) =0 $$

So , every solution of \( R^2(z) =z \) is asolution of \( R(z) =z \) .

Here comes the question of existence of periodic point .

I. N . Baker proved that ,

#### Theorem:

Let P be a polynomial of degree at least 2 and suppose that P has no periodic points of period n . Then n=2 and P is to \( z \rightarrow z^2 – z \) .

#### Theorem:

Let \( R , \ \ (\frac {P}{Q}) \) be a rational function of degree $$ d = max \{ degree(P) , degree(Q) \} , \ where \ d \geq 2 . $$