Is the Riemann zeta function surjective or does it miss one value?
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2$\begingroup$ I deleted my comment and posted an answer with all detail. $\endgroup$ – Alexandre Eremenko May 18 '18 at 2:50
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12$\begingroup$ Great, now we just need to get a better understanding of the preimage of $0$. $\endgroup$ – Mikhail Katz May 18 '18 at 8:52
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1$\begingroup$ Maybe of interest: en.wikipedia.org/wiki/Zeta_function_universality $\endgroup$ – Steve Huntsman May 18 '18 at 13:38
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4$\begingroup$ You should state that you initially asked this on Math SE with no answer. Link to thread there: Is the Riemann zeta function surjective? $\endgroup$ – Jeppe Stig Nielsen May 19 '18 at 9:22
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4$\begingroup$ By a result of Harald Bohr (brother of Niels the physicist), $\zeta(s)$ takes on any $z \in \Bbb{C}^\times$ as a value infinitely often, in any strip $1 < \Re(s) < 1+\varepsilon$ infinitely often (for a proof see sec 1.2 of mathematik.uni-wuerzburg.de/~steuding/univ-india-final.pdf). $\endgroup$ – Nicolas Schmidt May 19 '18 at 15:13
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The Riemann zeta function is surjective. First, $\zeta(1/z)$ is holomorphic in the punctured disk $0<|z|<1$. Looking at $z=(1/2+it)^{-1}$ with $t\to\infty$ reveals that $\zeta(1/z)$ has an essential singularity at $z=0$, hence $\zeta(s)$ misses at most one value. If $\zeta(s)=w$ then $\zeta(\overline{s})=\overline{w}$, hence $\zeta(s)$ can only miss a real value. However, $\zeta(s)$ maps the real interval $(1,\infty)$ onto $(1,\infty)$, and the real interval $(-2,1)$ onto $(-\infty,0)$. Also $\zeta(-19)>1$ and $\zeta(-18)=0$, hence $\zeta(s)$ maps $[-19,-18]$ onto a real interval that contains $[0,1]$. So $\zeta(s)$ does not miss any real value, and hence it does not miss any complex value.
answered May 18 '18 at 1:06
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$\zeta$ function has only one pole at $z=1$. It also has order $1$. If $\zeta$ omits $c\in C$ then $g:=1/(\zeta-c)$ is entire with one simple zero at $1$. As it is of order $1$, it must be $g(z)=(z-1)e^{az+b}$, by Hadamard's factorization theorem, so $$\zeta(z)=(z-1)^{-1}e^{-az-b}+c,$$ which is absurd.
The formula I wrote is the general form of a non-surjective meromorphic function of order $1$ with a single pole at $1$ and omitting $c$.
The reference on Hadamard's factorization theorem is any texbook on complex variables, for example Ahlfors, Complex Analysis, or Titchmarsh, Function Theory, or Whittaker Watson, or whatever you have.
answered May 18 '18 at 2:49
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1$\begingroup$ Now that we have two solutions, we can ask ourselves, Which is more elementary: Picard's Great theorem or Hadamard's factorization? $\endgroup$ – Gerald Edgar May 18 '18 at 11:57
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1$\begingroup$ @GeraldEdgar: Eremenko's proof is more to the point, and also more elementary, in my opinion. My proof is pretty ad hoc. $\endgroup$ – GH from MO May 18 '18 at 12:57
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4$\begingroup$ @Gerald Edgar: Picard's Great Theorem is probably more elementary then the general Hadamard's theorem (and it was proved earlier), but the special case of Hadamard's theorem that is used here is also quite simple. I only used the fact that an entire function without zeros is $e^g$, and has order one only if $g(z)=az+b$. $\endgroup$ – Alexandre Eremenko May 18 '18 at 13:19
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2$\begingroup$ This argument also shows that $\zeta$ takes any value infinitely many times. $\endgroup$ – Will Sawin May 20 '18 at 13:21
