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In the Ivić, Aleksandar (2013). The theory of Hardy's Z-function, page 16 the Hardy function is defined as

$$\begin{align} Z(t) &:= \zeta \left(\frac{1}{2}+it\right)\left(\chi\left(\frac{1}{2}+it\right) \right)^{-1/2}\\[2ex] &= \zeta \left(\frac{1}{2}+it\right)\left( \frac{\Gamma\left(\frac{1}{2}\,\left(1 - \left(\frac{1}{2}+it\right)\right) \right)}{\Gamma \left( \frac{1}{2}\,\left(\frac{1}{2}+it \right)\right)}\,\pi^{\left(\frac{1}{2}+it\right)-\frac{1}{2}} \right)^{-1/2}\\[2ex] &= \zeta \left(\frac{1}{2}+it\right)\left( \frac{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}\,\pi^{it} \right)^{-1/2}\\[2ex] &=\zeta \left(\frac{1}{2}+it\right)\left( \frac{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)} \right)^{-1/2}\; \,\pi^{\frac{-it}{2}}\\[2ex] &=\zeta \left(\frac{1}{2}+it\right)\color{blue}{ \frac{\Gamma^{1/2} \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma^{1/2}\left(\frac{1}{4} - \frac{it}{2} \right)}} \;\pi^{\frac{-it}{2}} \end{align}$$

The Hardy function is the same as the Riemann-Siegel Z function as the Wikipedia entry indicates. The definition in the Wikpedia entry is

$$ \begin{align} Z(t)&=\zeta\left(\frac{1}{2}+it\right)\; e^{i\theta(t)}\\[2ex] &=\zeta\left(\frac{1}{2}+it\right)\; e^{i\,\left[ \Im\left(\ln\Gamma\left(\frac{1}{4}+\frac{it}{2} \right) \right)-\frac{t}{2}\ln(\pi) \right]}\\[2ex] &=\zeta\left(\frac{1}{2}+it\right)\quad \color{blue}{e^{i \;\text{arg}\left(\Gamma(\frac{1}{4}+\frac{it}{2})\right)}}\quad \pi^{\frac{-it}{2}} \end{align}$$

So the question is (provided I haven't messed up the algebra), How do I see that

$$ \frac{\Gamma^{1/2} \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma^{1/2}\left(\frac{1}{4} - \frac{it}{2} \right)}=\;e^{i \;\text{arg}\left(\Gamma(\frac{1}{4}+\frac{it}{2})\right)}$$

?

I see that the argument of $z = x + i y=\rho\ e^{i\theta}$

$$\theta = \arg(z)= \frac{1}{i}\log\sqrt{\frac{z}{\bar z}}=\frac{\log z - \log \bar z}{2i}$$

so we can modify

$$\frac{\Gamma^{1/2} \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma^{1/2}\left(\frac{1}{4} - \frac{it}{2} \right)}= \sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}$$

but is $\Gamma\left(\frac{1}{4} - \frac{it}{2}\right)$ the complex conjugate of $\Gamma\left(\frac{1}{4} + \frac{it}{2}\right)$?

Antoni Parellada
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  • Set $;z:=\Gamma\left(\frac{1}{4}+\frac{it}{2}\right);$ then you want $\displaystyle \sqrt{\frac{z}{\overline{z}}}=\sqrt{\frac{z^2}{z\overline{z}}}=;e^{,i ;\text{arg}(z)};$ with $;\displaystyle e^{,i ;\text{arg}(z)}=\frac z{|z|};$... – Raymond Manzoni Jun 18 '17 at 11:01
  • It follows from the duplication formula @RaymondManzoni – reuns Jun 18 '17 at 15:44
  • Forget about the Z-function and look at the functional equation $$\Lambda(s) = \pi^{-s/2}\Gamma(s/2)\zeta(s) = \Lambda(1-s)$$ vs $$\lambda(s) = (X(s))^{1/2}\zeta(s) = \lambda(1-s)$$ – reuns Jun 18 '17 at 15:47
  • @Antoni Parellada: if your question is how to obtain the argument of $;\displaystyle \zeta\left(\frac{1}{2}+it\right);$ then see the "Quantitative discussion" from this answer up to $(4)$. – Raymond Manzoni Jun 18 '17 at 17:14
  • @RaymondManzoni Can you consider an answer explaining how $\displaystyle \sqrt{\frac{z}{\overline{z}}}=\sqrt{\frac{z^2}{z\overline{z}}}=;e^{,i ;\text{arg}(z)}$ follows from the duplication formula? – Antoni Parellada Jun 19 '17 at 02:37
  • @AntoniParellada: The duplication formula was proposed by user1952009 not by me. My point was precisely that no property of $\zeta$ or $\Gamma$ (except $,\overline{\Gamma(z)}=\Gamma(\overline{z}),$) was needed for your identity but only complex operations. At the left you have $;\displaystyle \sqrt{\frac{z}{\overline{z}}}=\sqrt{\frac{z,z}{z,\overline{z}}}=\frac {\sqrt{z^2}}{\sqrt{z\overline{z}}}=\frac {z}{|z|},$ (with $\sqrt{},$ the principal square root – Raymond Manzoni Jun 19 '17 at 12:24
  • i.e. $\displaystyle\sqrt{z}=\sqrt{|z|}e^{i\arg(z)/2},$ using with care the branch excluding the nonpositive reals). The usual $;\displaystyle \frac z{|z|}=e^{,i ;\text{arg}(z)};$ allows to conclude. Sorry if this was unclear, – Raymond Manzoni Jun 19 '17 at 12:24
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    From whichever definition of $\Gamma$ you use, it should be easy to see that $\Gamma(x) \in \mathbb{R}$ for positive real $x$. So the two entire meromorphic functions $f \colon z \mapsto \Gamma(z)$ and $g \colon z \mapsto \overline{\Gamma(\overline{z})}$ coincide on a set that has an accumulation point in the domain where they are holomorphic. By the identity theorem, $f \equiv g$. – Daniel Fischer Jun 19 '17 at 18:59
  • @DanielFischer OK. So the last sentence of the OP is therefore correct... Thank you. There is more than one unfamiliar concept in your comment, but it helps confirming the intuition behind the transition to the exponential form (unless I misunderstood you). – Antoni Parellada Jun 19 '17 at 19:02
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    @AntoniParellada You can think to the $Z(t)$ function like this : defined for $t$ real as $Z(t) = \zeta(1/2+it) e^{i \theta(t)}$ where $\theta(t)\in \mathbb{R}$ is chosen such that $Z(t)$ is real and analytic. It happens that we have a simple expression $\theta(t) = arg\ \Gamma(1/4+it/2) - \log(\pi)/2$ thanks to the functional equation. Thus $\zeta(1/2+it)$ has a zero at every sign change of $Z(t)$, and this is an important point in favor to the Riemann hypothesis – reuns Jun 19 '17 at 20:58

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