Poles of $f(z)$ when $f(z)$ has a zero and a singularity at the same point

Question

I want to determine the poles, and their orders, of $f(z)$.¹⁾

$$ f(z) = \frac{1+e^{i\pi z}}{(z-1)^2(z+1)^2} $$

The solution says that $f(z)$ has two simple poles at $z = +1$ and $z= -1$, but to me they both seem to be double poles? How can they be simple?

Maybe it has something to do with $f(z)$ having zeroes at $z=\pm 1$?

Answer 1

The function $g(z)=\frac z{z^2}$ has a simple pole at the origin. Your function does the same thing, only the functional expression hides it a little better.

Yes, it has everything to do with the numerator also having roots (of order 1) at those points. In general, if you have two meromorphic functions $f$ and $g$ and $f$ has a root of order $m$ at $a$ and $g$ has a root of order $n$, then $fg$ has a root of order $m+n$ (and $\frac fg$ has a root of order $m-n$). This includes poles (roots of negative order), and finite non-zero values (roots of order zero). You might have a removable singularity that needs removing before applying this.

Answer 2

The quotient function $f(z)=\frac{p(z)}{q(z)}$ has a pole of order $n$ at $z_0$ iff $q(z)$ has a zero of order $n$ at $z_0$ and $p(z)$ is analytic and nonzero at $z_0$ .

If $p(z)$ too has a zero of order $m<n$ at $z_0$ then $f(z)$ has a pole of order $n-m$ at $z_0$.