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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.

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

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  • $\begingroup$ 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

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