Let $\mathbb{C}_{an}$ be the expansion of the structure $(\mathbb{C}; +,-,×,0,1)$ by adding the restricted complex analytic functions. This is the complex analog of the familiar $\mathbb{R}_{an}$ in O-minimality. What do we know about the model theory of $\mathbb{C}_{an}$? Model-completeness? Quantifier-elimination? etc. PS. It seems Rückert's Nullstellensatz gives us model-completeness (This came to my mind just now) Is this true?
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The structure $\mathbb C_{an}$ is bi-interpretable with $\mathbb R_{an}$. To see this, take for example the restriction of the complex exponential function to the unit square S with corners 0,1,i,1+i. Then from this restriction you can define S as the set of points where the restricted function takes a non-zero value. Then the set of complex numbers $x$ satisfying $\forall y[y \in S \to xy \in S]$ is the unit interval [0,1]. Then taking $\pm$ and $1/x$ you can define $\mathbb R$.
So while many people do study restricted holomorphic functions, they do so usually in the context of o-minimality.
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2$\begingroup$ Welcome to [being active on] MathOverflow! $\endgroup$– Asaf Karagila ♦Commented Feb 23, 2018 at 14:55
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$\begingroup$ This answer relies on a rather accidental property that you chose to interpret "restricted analytic function" as "restricted to rectangles". It's also easily seen to work for e.g. disks, but what if we take just any odd shape? (Of course, if they shape is too odd, it will be no longer definable in $\mathbb R_{an}$, but I'm only concerned about the other direction.) More precisely: let $D$ be a bounded set with nonempty interior, and define $\mathbb C_{an,D}$ as the expansion of the complex field with restrictions to $D$ of all functions that are analytic in a neighbourhood of $\overline D$... $\endgroup$ Commented Feb 24, 2018 at 11:27
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$\begingroup$ ... Does $\mathbb C_{an,D}$ interpret $\mathbb R_{an}$ for every $D$? (One can check that without loss of generality, we may assume $D$ to be compact, regular closed (i.e., the closure of its own interior), and simply connected.) $\endgroup$ Commented Feb 24, 2018 at 11:30
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$\begingroup$ It does work if at least part of the boundary of $D$ is an analytic curve. $\endgroup$ Commented Feb 24, 2018 at 12:38