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I have $z\mapsto 1-1/z^2$ which has the periodic orbit {$1,0,\infty$} on the Riemann sphere. Next, I want to calculate the corresponding multiplier $\lambda= (f^{\circ n})' (z_i)=f'(z_1)\cdots f'(z_n)$. In this example we have $f'(z)=2/z^3$, hence $\lambda=f'(1) f'(0) f'(\infty)=2/1 \cdot 2/0 \cdot 2/\infty =\, ???$.

What is wrong here? (I shall prove that $\lambda=0$).

Thanks for your help!

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up vote 1 down vote accepted

Computing the derivative of $f^{\circ 3}(z)$ yield $$-\frac{8 z^3 \left(z^2-1\right)^3}{\left(2 z^2-1\right)^3}$$

Setting $z=1$ yields 0 in the numerator, and 1 in the denominator. Hence, $\lambda = 0.$

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