NOTE: This question is not about fractional derivatives of complex order. That topic has already been discussed on this site, here, for example. No - this question is more simple.
How do we define $\mathrm{D}^\nu f$ for $\nu\in\mathbb{C}\wedge\operatorname{Re}(\nu)>0$ when $f:\mathbb{C}\to\mathbb{C}$?
I am aware that for functions $f:\mathbb{R}\to\mathbb{R}$, the (Riemann-Liouville) derivative of order $\nu$ is defined as $$(\mathrm{D}^\nu f)(x)=\mathrm{D}^n\left\{\frac{1}{\Gamma(n-\nu)}\int_0^x(x-t)^{n-\nu-1}f(t)\mathrm{d}t\right\}$$ Here, $n=\left\lceil\operatorname{Re}(\nu)\right\rceil$. However, this definition seems incompatible with more general functions $f:\mathbb{C}\to\mathbb{C}$ - in particular, if $f$ is not path independent, the notation $$\int_0^z$$ Is pretty much meaningless, as discussed in my previous question. So, is there a "good" definition for fractional derivatives of complex argument functions? Which path of integration should we take?
MOTIVATION:
My ultimate goal is to try (I am not a priori convinced this is true) to prove that for functions $f:U\subseteq\Bbb{C}\to\Bbb{C}$ and $\nu,\alpha\in\mathbb{C}$, both having a positive real part, that the operator $\mathrm{D}$ is smooth in the $L^2$ sense, that is, $\forall\epsilon\in\mathbb{R}_{\geq 0}~,~\exists\delta\in\mathbb{R}_{\geq 0}$ such that $$|\nu-\alpha|<\delta\implies\left|\langle\mathrm{D}^\nu f,\mathrm{D}^\alpha f\rangle\right|=\left|\int_U \mathrm{D}^\nu f(z)~\overline{\mathrm{D}^\alpha f(z)}\mathrm{d}z\right|<\epsilon$$