# Extending holomorphic function on annulus to holomorphic function on disk

I'm stuck with the following problem and I'd appreciate a hint.

Suppose that $f$ is holomorphic on the annulus $r<|z|<R$. Show that if there exists some $\rho$ with $r<\rho<R$ such that $$\int_{|z|=\rho} z^nf(z)\,dz=0, \qquad \forall n\geq 0$$ then $f$ extends to a holomorphic function on $|z|<R$.

• Do you know the integral formula for the coefficients of the Laurent series of $f$? – Daniel Fischer Aug 7 '14 at 23:43
• Okay, I realize this was a really dumb question. Thank you. – user54631 Aug 7 '14 at 23:49
• No, it was not a dumb question. That the answer is easy if you know the right way does not mean a question is dumb. – Daniel Fischer Aug 7 '14 at 23:50
• A Prof I had years ago (a topologist) was a friend and colleague of Walter Rudin's wife (also a topologist.) This Prof was rather detailed and not inclined toward a bit of exaggeration, and he told a story about Walter. Near the beginning of a class, apparently Walter Rudin claimed that some fact was obvious, and someone asked why. Rudin became confused and left to go back to his office to check his notes. He didn't return until almost the end, and said, "Yes, it is obvious." Ah, the human mind. I think Walter's wife had told that story on him. :) – DisintegratingByParts Aug 8 '14 at 1:08

Since $f$ is holomorphic on the annulus $r<|z|<R$, f admits a Laurent expansion $$f(z)=\ldots+\frac{c_{-2}}{z^2}+\frac{c_{-1}}{z}+c_0+c_1z+c_2z^2+\ldots$$ where $$c_n=\frac{1}{2\pi i}\int_{|z|=\rho} \frac{f(z)}{z^{n+1}}\,dz, \qquad \forall r<\rho<R$$
By assumption $c_n=0$ for all negative $n$, so that $f$ is indeed holomorphic on the disk $∣z∣<R$.