Small debate between friends, how do you classify the singular points of the following function:
$$f(z) = \frac{z^3(z+1)}{1-z^4}$$
I found that the singular points are : $z=i, z=-i, z=1, z=-1$. Where both $z=i, z=-i, z=1$ are simple poles, $z=-1$ is a removable singularity.
However my friend says that $z=i ,z=-1$ are simple poles but $z=1, z=-1$ are poles of 2nd order.
Who is right or are we both wrong ?
Thank you!