I get the expansion of $h$ to be $$ h(z) = {1 \over z } \sum_{r=1}^{\infty}{1 \over r}{(-{\alpha \over z}})^r $$ $$ \Rightarrow h(z) = \sum_{r=-2}^{-\infty}{{(-\alpha)^{r+1} \over -(r+1)} z^{r}} $$ and thus the final integral to be $0$, since $ \int_{|z|=2} z^n dz = 0,$ since $2 \ge 1,$ the inside boundary of the annulus.
Is this right, or am I getting something mixed up?
Thanks! :)