On Ahlfors' Complex Analysis textbook, he shows a method to integrate integrals of the form $\int_0^\infty x^\alpha R(x)dx $ where $R(x) $ is rational function that has a zero of at least order two at infinity and at most a simple pole at the origin, and $0< \alpha < 1$. Letting $x=t^2$ the integral becomes $2 \int_0^\infty t^{2\alpha+1}R(t^2)dt$. If we define the branch of log to be between $-\frac{\pi}{2}$ and $\frac{3\pi}{2}$, i.e. omitting the negative imaginary axis, $z^{2\alpha}$ has its argument between $-\pi\alpha$ and $3\pi\alpha$. Now use a closed curve consisting of two line segments along the positive and negative axis and two semicircles in the upper half plane, one very large and one very small, both centered at the origin. Then $$\int_{-\infty}^\infty z^{2\alpha+1}R(z^2)dz = \int_0^\infty (z^{2\alpha+1}+(-z)^{2\alpha+1})R(z^2)dz $$ The textbook says that since "$(-z)^{2\alpha+1} = e^{2\pi i\alpha}z^{2\alpha}$", the integral equals $(1+e^{2\pi i \alpha})\int_0^\infty z^{2\alpha+1} R (z^2)dt $.
If arg $z$ is between $-\pi/2$ and $\pi/2$ then we know that $\arg (-z) = \arg(z) + \pi$ so that $(-z)^{2\alpha+1} = e^{2\pi i\alpha}z^{2\alpha}$. But if $\arg z$ is between $\pi/2$ and $3\pi/2$ we have by our choice of the branch of $\log$ that $\arg (-z) = \arg(z) - \pi$ so that $(-z)^{2\alpha+1} = e^{-2\pi i\alpha}z^{2\alpha}$. Am I missing something here?
Ahlfor says then that the integral can be computed with the residue theorem, and claims that the residues of $z^{2\alpha+1} R (z^2)$ are same as the residues of $z^\alpha R(z)$ in the whole plane.
Why are the residues of $z^{2\alpha+1} R (z^2)$ same as the residues of $z^\alpha R(z)$ in the whole plane?