Suppose $p(z)$ is a polynomial of even degree, all of whose roots satisfy $|z|<R$. Explain why there is a holomorphic (i.e. analytic) function $h(z)$ defined on the region $R<|z|<\infty$ which satisfies $h(z)^{2}=p(z)$. We write $h(z)=\sqrt{p(z)}$

By expanding in a Laurent series or otherwise, evaluate

$$\int_{C} \sqrt{z^{4}-z} d z$$

where $C$ is the circle of radius 2 with the anticlockwise orientation. (Your answer will be well-defined up to a factor of $\pm 1$, depending on which square root you pick.)