Can derivatives have simple poles?

Question

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?

Answer 1

You are correct, and it follows from the Laurent series expansion of $f$ as you noted.

Another way to see it is that a function with a simple pole at $z_0$ has a non-zero residue there, and integrals of derivatives are $0$ (because of the complex version of the fundamental theorem of calculus).

Answer 2

Suppose that $f'(z)= \frac{a}{z-z_0}+g(z)$ for $z \in D'(z_0,r)$ with a function $g$ which is holomorphic on $D(z_0,r)$ and $a \ne 0.$

Let $0<R< r$, then $\int_{|z-z_0|=R}f'(z) dz= a 2 \pi i \ne 0$.

On the other hand, since $f$ is an anti - derivative of $f'$ on $D'(z_0,r)$ and we integrate over a closed curve we have $\int_{|z-z_0|=R}f'(z) dz=0.$

A contradiction.

Answer 3

No:

• if $z_0$ is a removable singularity of $f$, then it is also a removable singularity of $f'$;
• if $z_0$ is a pole of order $n$ of $f$ ($n\in\mathbb N$), then it is a pole of order $n+1$ of $f'$, and therefore it will not be a single pole;
• if $z_0$ is an essencial singularity of $f$, then it is also an essencial singularity of $f'$.

Answer 4

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 $z_0$.