TITLE: Can derivatives have simple poles?
QUESTION [6 upvotes]: Let $f$ be be holomorphic on the punctured disk $D'(z_0,r)$ with $z_0$ being an isolated singularity. Is it possible for $f'$ to have a simple pole at $z_0$? Just looking at the Laurent expansion of $f$ about $z_0$ I think the answer should be "No". I am somewhat unsure about this. Can anyone help me out?

REPLY [2 votes]: No. Note that "suitably near" a simple pole at $z_0$, we have
$$f(z) \approx \frac{c}{z - z_0}$$
in effect, by definition. That would suggest that if $f$ is the derivative of some function $F$, then we would have
$$F(z) \approx \int \frac{c}{z - z_0}\ dz$$
around that point. But this is just the logarithm integral, and as we know, the logarithm has a branch point, not a normal singularity, as its "singularity", here located at $z_0$.