Riemann zeta function and modulus

Question

The functional equation for the zeta function $ζ(s)$ is given by

$ζ(s)=f(s)ζ(1-s)$

(a) We know that if $Re(s)=1/2$, then $|f(s)|=1$.

My question is about the case where $|f(s)|=1$ outside the critical line. Is this case possible, i.e.,

(b) Is this implication "if $|f(s)|=1$ then $Re(s)=1/2$" correct?

Answer 1

• (a) if $Re(s)=1/2$, then $|f(s)|=1$ : this is true and allows us to find a simple expression of the phase of the Riemann zeta function on the critical line.

• (b) the function $f(s)$ itself doesn't contain $\zeta$ since $f(s)$ is, from the functional equation : $$\tag{1}f(s)=2^s\pi^{s-1}\sin\left(\frac {\pi s}2\right)\Gamma(1-s)$$ but $|f(s)|$ may be $1$ out of the critical line as you may see on this picture of $|f(x+i y)]-1$ :

  The vertical line at $x=\frac 12$ gives the solution $(a)$ but two solutions to $|f(x+i y)]=1$ exist for $y$ near $2\pi$ and $-2\pi$ (for $x$ near $\frac 12$ see too this discussion concerning Siegel-$\theta$). You'll have to exclude these two solutions to make your implication correct!

$\qquad$Another perspective view :

And with this picture for $x$ between $-30$ and $30$ a kind of symbol for your 'superquest' :-)

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APPROXIMATIONS:
In the remaining we will always suppose that $x\in[0,1]$ and will only consider the case $y>0$.

Let's remember $6.1.30$ from A&S : $$\tag{2}\left|\Gamma\left(\frac 12+iy\right)\right|^2=\frac {\pi}{\cosh(\pi y)}$$ An excellent approximation for other values of $x$ (when $y>2$) is given by : $$\tag{3}\left|\Gamma\left(x+iy\right)\right|^2\approx\frac {\pi\;y^{2x-1}}{\cosh(\pi y)}$$ (the error is less than $0.4\%$ and quickly decreasing with $y$)

Since $\ \left|\sin\left(\frac {\pi (x+iy)}2\right)\right|^2=\cosh^2\bigl(\frac{\pi\;y}2\bigr)-\cos^2\bigl(\frac {\pi\;x}2\bigr)\ $ and $\ \left|(2\pi)^{x+iy}\right|^2=(2\pi)^{2x}\;$ we get :

\begin{align} |f(x+iy)|^2&\approx\left|\frac{(2\pi)^{x+iy}}{\pi}\sin\left(\frac {\pi (x+iy)}2\right)\right|^2\frac {\pi\;y^{1-2x}}{\cosh(\pi y)}\\ &\approx\frac{(2\pi)^{2x}}{\pi}\left(\cosh^2\left(\frac{\pi\;y}2\right)-\cos^2\left(\frac {\pi\;x}2\right)\right)\frac {\;y^{1-2x}}{\cosh(\pi y)}\\ \tag{4}&\approx \left(\frac{2\pi}y\right)^{2x}\frac y{\pi}\frac{\cosh^2\left(\frac{\pi\;y}2\right)-\cos^2\left(\frac {\pi\;x}2\right)}{\cosh(\pi y)}\\ \end{align}

Now for $y\gg 1$ the fraction at the right will converge to $\frac 12$ (since $|\cos|\le 1$) and we will have : $$\tag{5}|f(x+iy)|\sim \left(\frac{2\pi}y\right)^{x-1/2}\quad\text{for}\ y\gg 1$$ which shows clearly all the things of interest for us :

• for $x=\frac 12$ we get $|f(x+iy)|=1$ independently of $y$
• there is only one other solution : $y\approx 2\pi$ (considering $y>0$)
• the visual aspect of the approximation shows no real difference :

Of course this is not a complete proof : the error term in $(3)$ to next order should be found (possibly using A&S' expansions and propagated) proving that $x\mapsto |f(x+iy)|$ is decreasing for $y>K$ and increasing for $y<K$ (with $K$ some constant near $2\pi$). $|f(x+iy)|$ should be studied with more care for $y$ near $0$ and so on. But, at least in principle, it appears ok to me so : Fine continuation!