$2\pi i\beta = \int\limits_{|z|=r} \frac{dz}{f(z)-z}$

If $\beta |z|^k < 1$ then $$ \frac{1}{f(z)-z} = \frac{1}{z^{k+1}(1-\beta z^k)} $$ can be developed into a Laurent series: $$ \frac{1}{z^{k+1}}(1 + \beta z^k + \beta^2 z^{2k} + \ldots) = \frac{1}{z^{k+1}} + \frac{\beta}{z} + \beta^2 z^{k-1} + \ldots $$ and the residuum at $z=0$ is the coefficient of $z^{-1}$: $$ \frac{1}{2 \pi i}\int\limits_{|z|=r} \frac{dz}{f(z)-z} = \operatorname{Res}(\frac{1}{f(z)-z}, 0) = \beta \, . $$