Let $C$ be the circle $|z-z_0| = r$ traversed counter-clockwise, and let $\alpha$ ne any nonzero real number. Parameterize $C$ by $z=z_0+re^{i\theta}$, with $-\pi < \theta < \pi$, and compute that $$\int_C(z-z_0)^{\alpha-1}dz = 2iR^{\alpha}\frac{sin(\pi \alpha)}{\alpha}$$ on the principal branch of the integrand.
I used the paramaterization to get that
$$
\int_C(z-z_0)^{\alpha -1}dz = \int_{-\pi}^{\pi}(z_0+re^{i\theta}-z_0)^{\alpha-1}d\theta = \int_{-\pi}^\pi(re^{i\theta})^{\alpha-1}d\theta.
$$
Carrying out this integral gives me
$$
\frac{-ir(e^{i\theta})^{\alpha-1}}{\alpha-1}|_{-\pi}^{\pi}.
$$
If I evaluate this, I get nowhere near the RHS of the original integral.
For one thing you didn't replace $dz$. You have $dz=rie^{i\theta}d\theta$. You end up with
$$ir^{\alpha}\int_{-\pi}^{\pi} e^{i\alpha\theta} d\theta =ir^{\alpha}\dfrac{e^{i\alpha\theta}}{i\alpha}\Big|_{-\pi}^{\pi}$$
Evaluate, then use $e^{it}=\cos t+i\sin t$.