Is the Riemann zeta function surjective? Is the Riemann zeta function surjective or does it miss one value?
 A: 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.
A: (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 $f(z) = a + |a| (z-3/4)$.
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 zeta-like 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 probably classical, but I first saw it in a paper on simple $a$-points by Gonek, Lester and Milinovich.
A: $\zeta$ 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.
