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

#### 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 .$$ (1) \Rightarrow d^{N-1} \leq N \\ therefore , N= 1 + (N-1) \leq 1 + (N – 1)(d – 1) \leq [1 + (d-1)]^{N-1} = d^{N – 1} \leq N $$