Does the function $\zeta(s)$ attain all complex values?

  • $\begingroup$ The Little Picard Theorem does not apply because $\zeta$ is not entire. $\endgroup$
    – Shimrod
    May 16, 2018 at 13:07
  • $\begingroup$ Still $(s-1)\zeta(s)$ is an entire function. Also, necessarily $\zeta(s)$ has an essential singularity in $\infty$, so I guess $\zeta(s)$ attains every complex value except possibly one in every (punctured) neighborhood of $\infty$? $\endgroup$ May 16, 2018 at 13:27
  • $\begingroup$ What about that one value? $\endgroup$
    – Shimrod
    May 16, 2018 at 13:28
  • 3
    $\begingroup$ You should state that you later asked the same question on Math Overflow. And got an answer! Link to thread there: Is the Riemann zeta function surjective? $\endgroup$ May 19, 2018 at 9:24


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