What Holomorphic?

From the previous post "Clock Face And Life, Two Pies And Habits" dated 30 Jul 2015,

$\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }=-\int^{\infty}_{-\infty}{it.g(\omega t)e^{-i\omega t}}d\,\omega=-it.G(t)$

$\theta=\omega t$ and $z=R_\theta e^{i\theta}$

$R_\theta$ being a function of $\theta$.

$\int { f(z)d(e^{ -i\theta  }) }$

takes a free ride with Fourier.  Must it still be holomorphic?

$f(z)=g(\omega t)$

as long as the Fourier tansform of $g(\omega t)$ exist,  $\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta  }) }$ is valid.  In the case of simple periodicity, $T$ a constant,

$\int ^{\infty}_{-\infty}{ f(z)d(e^{ -i\theta  }) }=\int ^{T}_{0}{ f(z)d(e^{ -i\theta  }) }=-it.G(t)|^{T}_0=-i.TG(T)$

where $G(t)$ is the Fourier transform of $f(z)$ with $z$ replaced by $\omega$.  $-i$ is in the direction of $\theta=0$.

Most liberating.

Note: $-i.tG(t)$ is a complex function ($\theta=\omega t$ being real), with a dual $f(z)$, running around in circles in the complex plane.

$e^{i\theta}$ being replaced with $e^{-i\theta}$.