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$$P^m(z) = z \ and \ P^N(z)=z \ where \ m|N \Rightarrow (P^m(z) – z) | (P^N(z)-z)$$

#### The proof of the theorem in Part 0 :

Let , P be the polynomials satisfying the hypothesis of theorem 6.2.1 .

Let , $$K = \{ z \in C | P^N(z) =z \} \\$$

and let $$M =\{ m \in Z : 1 \leq m \leq N , m|N \} \\$$

then each $$z \in K$$ is a fixed point of $$P^m$$ for some $$m \in M$$ and we let m(z) be the minimal such m . $$\\$$

The proof depends on establishing the inequalities , $$d^{N-1} (d-1) \leq \sum_{k} [\mu (N, z) – \mu (m(z) ,z) ] \\ \leq N(d-1)$$ ………………………………..$$(1)$$ where $$\mu (n, w)$$ is the no of fixed points of $$P^n$$ at w .$$(1) \Rightarrow d^{N-1} \leq N \\ therefore , N= 1 + (N-1) \leq 1 + (N – 1)(d – 1) \leq [1 + (d-1)]^{N-1} = d^{N – 1} \leq N$$