Inverse Z-transform with non-integer exponent

Question

I would like to find the inverse Z-transform of $$H(z)=1-z^{\alpha},$$ where $\alpha$ is a real but not integer number. Is there any formulation that I can use? I think the way to do it is by contour integral but I don't know how to use the method on this function.

Answer 1

The $Z$-transform $$H(z) = \sum_n h(n) z^n$$ is well-defined iff there is $r> 0, \epsilon >0$ where it is converges for $|z| \in (r-\epsilon,r+\epsilon)$. In that case $H(z)$ is analytic on this annulus we can define the inverse $$h(n) = \frac{1}{2i\pi} \int_{|z|=r} \frac{h(z)}{z^{n+1}}dz = r^{-n}\int_0^1 h(r e^{2i \pi t}) e^{-2 i \pi n t}dt$$ Here for $\alpha \not \in \mathbb{Z}$, $H(z) = 1-z^a$ is not analytic on any such annulus, thus by definition it is not a Z-transform.

Now if you look at $H(e^{2i \pi t}) =1 - e^{2i \pi \alpha}$ then you can compute its Fourier coefficients $$h(n) = \int_0^1 h( e^{2i \pi t}) e^{-2 i \pi n t}dt = 1_{n=0}- \frac{e^{2 i \pi (\alpha-n)}-1}{2i \pi (\alpha-n)}$$ Obtaining a Fourier series $$H(e^{2i \pi t}) = \sum_{n=-\infty}^\infty h(n) e^{2 i\pi nt}$$ which converges for $t \ne 0$