# Contour integral of $\frac{1}{\sqrt z}$ with branch cut

I am a physicist who usually doesn't need to care about the fact that square root is not single-valued on the complex plane. But I would like to give a meaning to and compute the contour integrals :

\begin{align} \oint_w \frac{z-a}{(z-w)^{1/2}}dz \end{align} and \begin{align} \oint_w \frac{z-a}{(z-w)^{3/2}}dz, \end{align} where $a$ is a complex number and the contour is a small circle around $w$. I have some very basic knowledge of complex analysis, know how to use residues, but I am not familiar with contour integrals which branch cuts. I don't find people considering these integrals online, so I would like to know if it is possible to give them a meaning and to compute them.

For the first integral, if the variable is changed so that the contour is around 0, there is a branch cut from $-\infty$ to 0. For the second integral there would be two branch points, $w$ and $-w$. I can deform the contours to avoid branch cuts, but how can I then compute the different integrals?

To evaluate the contour integrals, you can simply parameterize. For the first example, set $z=w+r e^{i \phi}$, $\phi \in [-\pi,\pi]$. Then the integral is equal to
$$r^{1/2} \int_{-\pi}^{\pi} d\phi \, e^{i \phi/2} \left (w-a+r e^{i \phi}\right ) = 4 (w-a) r^{1/2} - \frac{4}{3} r^{3/2}$$