Let $f$ be analytic in $D(0,2)$. Assume that for all $n∈\mathbb{N}$ $\int_{ |z|=1} {f(z)\over(n+1)z−1}dz=0$. Prove that $f(z)=0$ for all $z∈D(0,2)$.

I'm thinking about a contradiction proof. Assuming that $f(z)$ doesn't not equal zero and contradicting the assumption. I'm not sure though. Any solutions or hints are greatly appreciated.


Hint: Look what is it $f(\frac{1}{n+1})$ by Cauchy's integral formula.

  • $\begingroup$ Thank you for responding. I'm not sure I understand the hint though. $\endgroup$ – Happy Apr 25 '16 at 17:39
  • $\begingroup$ @DerpMagoo Do you know what is it Cauchy's integral formula? $\endgroup$ – Cortizol Apr 25 '16 at 17:41
  • $\begingroup$ Yes, this, right? en.wikipedia.org/wiki/Cauchy%27s_integral_formula $\endgroup$ – Happy Apr 25 '16 at 17:45
  • $\begingroup$ @DerpMagoo Yes. And if you take $a=\frac{1}{n+1}$ and look at your condition, you will get... Try it. $\endgroup$ – Cortizol Apr 25 '16 at 17:48
  • $\begingroup$ Since $f$ is analytic I can choose $a={1\over (n+1)}$ and substitute $a$ into the cauchy integral formula which gives me $0$ from the condition and thus $f=0$? $\endgroup$ – Happy Apr 25 '16 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.