Cauchy differentiation formula says $$f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_{\gamma} \frac{f(z)}{(z-z_0)^{n+1}}dz $$
Would changing the $n$ by $\alpha\in \mathbb{R}$ and factorial by $\Gamma$ make it a fractional derivative, i.e. $$f^{(\alpha)}(z_0)=\frac{\Gamma(\alpha+1)}{2\pi i}\int_{\gamma} \frac{f(z)}{(z-z_0)^{\alpha+1}}dz $$
This way or another, is the following correct or how to do it correctly $$\frac{1}{2\pi i}\int_{|z|=2} \frac{\frac{1}{z-1}}{\sqrt{z-z_0}}dz = \frac{1}{\Gamma(-1/2+1)}\frac{d^{-1/2}}{dz^{-1/2}}\frac{1}{z-1}\Big|_{z=z_0} =-\frac{i}{\sqrt{z_0-1}}$$
Edit The question above come from trying to find, for some numerical experiment, a function $x(t)$ (that will be an exact solution of somehting) satisfying $$\frac{1}{2\pi i}\int_\gamma x(t)t^{-k-1} dt=0, \quad k=0,1,2,3...K$$ (K some finite number) such that $$\frac{1}{2\pi i}\int_{\gamma} \frac{x(z)}{(z-z_0)^{\beta}}dz $$ is analytically known/obtainable (in specific branch if needed). Consider $\gamma$ is something regular and smooth, circle or ellipse. $\beta=1/2$ was choose for simplicity, i.e. can be anything else.
The contour integral is not uniquely defined if $\alpha$ is fractional, because $1/(z-z_0)^{\alpha+1}$ is multi-valued and the contour crosses a branch cut.