Is the Riemann zeta function surjective or does it miss one value?

3$\begingroup$ I deleted my comment and posted an answer with all detail. $\endgroup$– Alexandre EremenkoCommented May 18, 2018 at 2:50

18$\begingroup$ Great, now we just need to get a better understanding of the preimage of $0$. $\endgroup$– Mikhail KatzCommented May 18, 2018 at 8:52

3$\begingroup$ Maybe of interest: en.wikipedia.org/wiki/Zeta_function_universality $\endgroup$– Steve HuntsmanCommented May 18, 2018 at 13:38

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 NielsenCommented May 19, 2018 at 9:22

6$\begingroup$ 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.uniwuerzburg.de/~steuding/univindiafinal.pdf). $\endgroup$– Nicolas SchmidtCommented May 19, 2018 at 15:13
3 Answers
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.
$\zeta$ function has only one pole at $z=1$. It also has order $1$. If $\zeta$ omits $c\in C$ then $g:=1/(\zetac)$ is entire with one simple zero at $1$. As it is of order $1$, it must be $g(z)=(z1)e^{az+b}$, by Hadamard's factorization theorem, so $$\zeta(z)=(z1)^{1}e^{azb}+c,$$ which is absurd.
The formula I wrote is the general form of a nonsurjective 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.

1$\begingroup$ Now that we have two solutions, we can ask ourselves, Which is more elementary: Picard's Great theorem or Hadamard's factorization? $\endgroup$ Commented May 18, 2018 at 11:57

2$\begingroup$ @GeraldEdgar: Eremenko's proof is more to the point, and also more elementary, in my opinion. My proof is pretty ad hoc. $\endgroup$ Commented May 18, 2018 at 12:57

5$\begingroup$ @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 $g(z)=az+b$. $\endgroup$ Commented May 18, 2018 at 13:19

4$\begingroup$ This argument also shows that $\zeta$ takes any value infinitely many times. $\endgroup$ Commented May 20, 2018 at 13:21
(Just a quick aside: the set of solutions to the equation $\zeta(s) = a$ when $a \neq 0$ are usually called the $a$points of the Riemann zeta function in the literature, in case you want to look up the state of the art)
As Steve Huntsman mentions in a comment, it is indeed possible to use the universality of the zeta function to prove that $\zeta(s)$ is surjective with a simple argument based on Rouché's theorem.
In particular, let $U$ be a compact connected subset with a smooth boundary in the critical strip on which universality holds (for simplicity, take $U = \{ z \in \mathbb{C} : z  3/4 \leqslant 1/8 \}$), and let $f:U \to \mathbb{C}$ be a function holomorphic on $U$ (i.e. holomorphic in the interior, and continuous on the boundary) with the following properties:
$f$ has no zeroes in $U$.
There exists a $z \in U$ such that $f(z) = a$.
For all $z \in \partial U$, $f(z) \neq a$.
As an example, for $U$ as above, one could take $f(z) = a + a (z3/4)$.
Now note that 3 implies that there is an $\epsilon$ such that $$0 < \epsilon < \inf_{z \in \partial U} f(z)  a.$$
Then, due to 1, universality guarantees that there is a $t \geqslant 0$ such that $$ \zeta(s+it)  f(s) < \epsilon$$ for every $s \in U$. In particular, this inequality holds for every $s \in \partial U$. Thus, for every $s \in \partial U$, we have that
$$ \zeta(s+it)  f(s) < \epsilon < \inf_{z\in\partial U} f(z)  a \leqslant f(s)  a. $$
Thus, Rouché's theorem implies that $f(s)  a$ and $f(s)  a + \zeta(s+it)  f(s) = \zeta(s+it)  a$ have the same number of roots in the interior of $U$. Since $f$ has such a root, it follows that $\zeta$ has such a root as well, and we are done.
This argument is robust in a few ways. For one, it applies to other functions which exhibit universality (which is known for fairly large classes of $L$functions and other zetalike families of interest). For another, effective versions of Voronin's universality theorem can be used to give a good lower bound for the number of $a$points in a rectangle within the critical strip. For a third, this argument actually proves that there are simple $a$points for every $a \neq 0$.
This argument is from a paper on simple $a$points by Gonek, Lester and Milinovich.