Consider the set $S=\{re^{i\theta}:r>0,-\pi<\theta<\pi\}$, i.e just $\mathbb C$ without a branch cut. Let $T\in B(X)$ for some Banach space $X$ with $\sigma(T)\subset S$, and let $\alpha,\beta\in (0,1]$. I want to use the definition of the Holomorphic Functional Calculus (H.F.C) found in Rudin to prove that $T^{\alpha\beta}=(T^\alpha)^\beta$. Now I am aware that Watanabe proved this result long ago, but in doing so he used Kato's definition of fractional powers, which I am unfamiliar with and would like to avoid (I'm sure it is most probably equivalent to the way I want to do it, but I'd prefer baby steps. I'm also sure the way I want to do it has been done before, but I can't find it.)
So how I would like to proceed is the following: Via the H.F.C we can define: $$T^\alpha=\frac{1}{2\pi i}\int_\Gamma z^\alpha(zI-T)^{-1}dz,$$ where $\Gamma$ is any contour surrounding $\sigma(T)$ and contained in $S$, because $z\mapsto z^\alpha$ is holomorphic on $S$. We now note that the Spectral Mapping Theorem, along with our assumption on $\alpha$, guarantees that $\Gamma$ will also surround $\sigma(T^\alpha)$. Thus we can define $$(T^\alpha)^\beta=\frac{1}{2\pi i}\int_\Gamma z^\beta(zI-T^\alpha)^{-1}dz.$$ Finally, we can also define $$T^{\alpha\beta}=\frac{1}{2\pi i}\int_\Gamma z^{\alpha\beta}(zI-T)^{-1}dz.$$
So in conclusion, what I want to prove is that: $$\frac{1}{2\pi i}\int_\Gamma z^{\alpha\beta}(zI-T)^{-1}dz=\frac{1}{2\pi i}\int_\Gamma z^\beta\left(zI-\left(\frac{1}{2\pi i}\int_\Gamma \lambda^\alpha(\lambda I-T)^{-1}d\lambda\right)\right)^{-1}dz,$$ using only the basic properties of the H.F.C. Unfortunately, I cannot untangle the right side of the equality, most probably because my complex analysis skills are not what they should be. I'd appreciate any help in this matter.
If you choose the contours carefully, I think this should work: \begin{align} T^{\alpha} & = \frac{1}{2\pi i}\oint_{C}z^{\alpha}(zI-T)^{-1}dz \\ (zI-T^{\alpha})^{-1} & =\frac{1}{2\pi i}\oint_{C}(z-w^{\alpha})^{-1}(wI-T)^{-1}dw \\ (T^{\alpha})^{\beta} & =\frac{1}{2\pi i}\oint_{C'}z^{\beta}(zI-T^{\alpha})^{-1}dz \\ & = \frac{1}{2\pi i}\oint_{C'}z^{\beta}\frac{1}{2\pi i}\int_{C}(z-w^{\alpha})^{-1}(wI-T)^{-1}dwdz \\ & = \frac{1}{2\pi i}\int_{C}\frac{1}{2\pi i}\oint_{C'}(z-w^{\alpha})^{-1}z^{\beta}dz (wI-T)^{-1}dw \\ & = \frac{1}{2\pi i}\int_{C}w^{\alpha\beta}(w I-T)^{-1}dw= T^{\alpha\beta}. \end{align}