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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 $$d = max \{ degree(P) , degree(Q) \} , \ where \ d \geq 2 .$$