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Let us consider the following function: $\phi(s)=\phi_1(s)+i\phi_2(s)$, given uniquely by the polar form $\phi(s)=\rho(s)\exp(i\theta(s))$, where $$\rho(s)=\sqrt{\phi_1^2(s)+\phi_2^2(s)}\neq 0$$ and $\theta(s)=\arg\phi(s)\in\mathbb{R}$, is the argument of $\phi(s)$.

My question is: What happen if the argument $\theta(s)=\arg\phi(s)\in\mathbb{R}$ is zero for $s$ in the set $D=\{s=\alpha+i\beta\in\mathbb{C}, 0<\alpha<1\}$. I know about the analytic continuation. But I am not sure if this principle would be applied here just like the case of modulus.

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    Um, if the argument is zero, then $\phi(s)$ is always a positive real. Which kind of thing do you expect to "happen" then? (Among the things we can conclude is that $\phi$ is either a constant or not analytic).2012-12-22
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    How I can make the difference between the two cases in your claim: "Among the things we can conclude is that $ϕ$ is either a constant or not analytic". The function $ϕ(s)$ is given by: $2(((1-2^{-1+s}))/((1-2^{s})))π^{s-1}sin(((πs)/2))Γ(1-s)$, where $Γ(1-s)$ is the gamma function.2012-12-22
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    @user53124: Since that formula is obviously analytic (and isn't constant either, because otherwise you could derive a closed expression for the gamma function), it cannot possibly have zero argument everywhere in _any_ open subset of its domain.2012-12-22
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    Yes, thank you very much. You make things very clearer.2012-12-22

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Since $ϕ(s)$ is an non-constant analytic function, then it cannot possibly have zero argument everywhere in any open subset of its domain. If so, then a closed expression for the gamma function can be obtained and this is impossible. However, this is possible in some regions.