Define a branch of $\frac{z^\alpha}{z^2+1}$ Define a branch of  $\frac{z^\alpha}{z^2+1}$. $\alpha$ is considered real and in the interval $(-1,1)$
Sketch the branch cut and the poles in the complex plane.
I have that the poles are $z=i, z=-i$. How do I define the branch? Would it run between $(-1,1)$? If so, can you explain why. If it is somewhere else, why?
 A: You need to define a branch whenever $\alpha$ is not an integer, because then $z^\alpha$ is a multi-valued function. A branch means consistently choosing one of the possible values.
To see why $z^\alpha$ is multivalued, write
$$z^\alpha=e^{\alpha \log z}$$
The complex logarithm $\log z$ is a multi-valued function. Representing $z=re^{i\phi}$ in polar coordinates, i.e. $r>0$ and $\phi\in(-\pi,\pi]$, $\log z$ could mean any of the values $$\log r + i(\phi+2\pi k)$$
where $k$ runs through the integers. This is because $e^{2\pi i k}=1$ for all $k\in\mathbb{Z}$.
Therefore you need to choose a branch of the logarithm to give your expression a meaning as a function. The principal branch of the logarithm is for example given by
$$\log z = \log r + i\phi$$
This is a holomorphic function on the complex plane without the negative real axis, i.e. on $\Omega=\mathbb{C}\backslash\{x\in\mathbb{R}, x\le 0\}$.
The negative real axis is called the branch cut. It is determined by the choice of the interval $(-\pi,\pi]$ in which we said our $\phi$ should lie and by our choice of $k=0$. Had we chosen a different interval, we would get the branch cut at a different angle.
Denoting by $l:\Omega\rightarrow\mathbb{C}$ the principal branch of the logarithm defined as above, we can define a branch of your function $\frac{z^\alpha}{z^2+1}$ by
$$\frac{e^{\alpha l(z)}}{z^2+1}$$
which is a meromorphic function on $\Omega$.
