What can we say about $f$ if $\int_{|z|=1}\frac{f(z)}{((k+1)z-1)^n}dz=0$?

Question

Consider $f$ an entire function that $\int_{|z|=1}\frac{f(z)}{((k+1)z-1)^n}dz=0$ for any $k\in\Bbb N$. What can we say about $f$ for $n\in\{ 1,2,3,\cdots\}$?

We know that $\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(z)}{((k+1)z-1)^{n+1}}dz = \frac{1}{(k+1)^{n+1}} f^{(n)}\left(\frac1{k+1}\right)$. Then $f^{(n)}\left(\frac1{k+1}\right) = 0$ for all $n\in\Bbb N$. Since $f$ is entire, can we say that $f^{(n)}(z) = 0$ for all $z\in\Bbb C$?

And if this is true, what we can say about $f$ is that it is polynomic of degree $n-1$ in each case?

If not, what else can I do?

Answer 1

Since $f$ is entire, we deduce that $G=f^{(n)}$ is entire to. In particular $$ G(\frac{1}{k+1})=0 \qquad k\ge 1. $$ and hence, $G(0)=G(\lim_k\frac{1}{k+1})=\lim_k G(\frac{1}{k+1})= 0$, by Analytic continuation, $G(z)=0$. Since $G=f^{(n)}=0$, we deduice that $f$ is a polynomial of degree at most $n-1$.

Answer 2

Consider the set $$D = \left\{\frac{1}{k+1}: k \in \mathbb{N}\right\}$$ Since $f^{(n)}|_D\equiv 0$, $f^{(n)}$ is holomorphic and $0$ is an accumulation point of $D$, frοm the identity theorem we conclude that $f^{(n)}\equiv 0$ on $\mathbb{C}$. From that it's easy to conclude that $f$ is a polynomial of degree at most $n-1$.