Does the function $\zeta(s)$ attain all complex values?
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$\begingroup$ The Little Picard Theorem does not apply because $\zeta$ is not entire. $\endgroup$– ShimrodMay 16, 2018 at 13:07
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$\begingroup$ 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 $\infty$? $\endgroup$– Jeppe Stig NielsenMay 16, 2018 at 13:27
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$\begingroup$ What about that one value? $\endgroup$– ShimrodMay 16, 2018 at 13:28
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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? $\endgroup$– Jeppe Stig NielsenMay 19, 2018 at 9:24
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