Let $f$ holomorphic at $D=\{z:0<|z|< 1\}$ Prove that exist a unique $c\in\mathbb{C}$ s.t $f(z)-\frac{c}{z}$ has primitive function.

Question

Let $f$ holomorphic at $D=\{z:0< |z|< 1\}$ Prove that exist a unique $c\in\mathbb{C}$ s.t $f(z)-\frac{c}{z}$ has primitive function in D.

So my thoughts were that we can write a laurent series for $f$ around $0$ and if $0$ is a simple pole the question is trivial because and $c$ is the laruent coefficent of $a_{n-1}$ and it's unique.
If $z=0$ is not a simple pole i'm not sure how to prove this.
Any hints will be welcome

Answer 1

If the Laurent series is $\sum_{n \in \mathbb Z} a_n z^{n}$ then $f(z)-\frac {a_{-1}} z $ is the derivative of $\sum_{n \in \mathbb Z \setminus {\{-1\}}} a_n \frac {z^{n+1 }} {n+1}$. This series converges to a holomorphic function in $0<|z|<1$.

Uniqueness is obvious since $\frac {c_1-c_2} z$ has no pri mitive if $c_1 \neq c_2$.