The idea that $|z|^\alpha$ for $\alpha \in (-1,0)$ has a removable singularity seems somewhat draconian. But by the proof of Riemann's theorem,
since we have $\lim_{|z| \to 0} |z \times z^\alpha|=\lim_{r \to 0} |re^{i\theta} \times r^\alpha|=\lim_{r \to 0} r^{1-\alpha}=0$, it must be true that $|z|^\alpha$ has a removable singularity.
My aim in all this is to find the fourier transform of $|x|^\alpha$. If $|z|^\alpha$ has a removable singularity, then how is one supposed to use residue theorem to find the foureir transform of $|x|^\alpha$?