31
$\begingroup$

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

$\endgroup$
65
$\begingroup$

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.

$\endgroup$
57
$\begingroup$

$\zeta$ 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.

$\endgroup$
  • 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
  • 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
  • 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
  • 2
    $\begingroup$ This argument also shows that $\zeta$ takes any value infinitely many times. $\endgroup$ – Will Sawin May 20 '18 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.