I'm reading Conway's complex analysis book and on page 103 he said:


  1. Is it true the converse? If $\int_{\gamma}f=0$ for any closed curve $\gamma$ in the punctured disk, then $f$ has an analytic extension to $B(a,R)$?

  2. What the fact $\lim_{z\to a}$ exists has to do with everything?


1 Answer 1

  1. The converse is not true. Take for example $f(z) = \dfrac{1}{(z-a)^2}$ (or any other function admitting an antiderivative on the punctured disc).
  2. If there is an analytic continuation, then the limit exists (since holomorphic functions are continuous).
  • $\begingroup$ Maybe it's woth mentioning that the integral is just picking out the residue. So you just had to find a meromorphic function with zero residue. This also tells you how to fix the claim. One would need that $\displaystyle \int (z-a)^k f(z)\, dz=0$ for all $k\geqslant 0$. (this is for OP) $\endgroup$ Nov 23, 2016 at 12:40
  • $\begingroup$ Sure, if you like. (Note that having residue 0 at every singularity is exactly the same as saying the function has an antiderivative on the complement of the singularity set). $\endgroup$
    – mrf
    Nov 23, 2016 at 12:41
  • $\begingroup$ Of course, it's just personally more apparent to me from the residue perspective. :) Maybe the OP will find some use out of the statement. $\endgroup$ Nov 23, 2016 at 12:42

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