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I am aware that a quasiconformal map satifies the formula $$ \frac{\partial f}{\partial \overline{z}} = \mu(z) \frac{\partial f}{\partial z} $$ where $\sup\{\mu(z):z \in \text{Domain}\{f\}\}<1$ imposes a bound on the eccentricity of the ellipses in the image of $f$. For a multicomplex function $F(z_1, z_2, \dots, z_n)$, one could impose the quasiconformality condition for each variable $z_k$. I have read computer science articles about three-dimensional quasiconformal mappings, but no explicit formula relating the conjugate partial derivative to the partial derivative is given.

Question: Does a generalization of the formula $\frac{\partial f}{\partial \overline{z}} = \mu(z) \frac{\partial f}{\partial z}$ apply to quasiconformal mappings in $\mathbb{R}^{2n+1}$?

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The correct definition of higher-dimensional quasiconformal maps does not use complex variables. The correct condition is $$ |Df(x)|\le K |J_f(x)| $$ where $J_f$ is the Jacobian determinant. An orientation-preserving homeomorphism $f$ between two domains $U, V$ in $R^n$ is called quasiconformal if it belongs to the Sobolev class $W^{n,1}_{loc}(U)$ (hence, has 1st order partial derivatives a.e. in $U$) and there exists a constant $K$ such that the above inequality is satisfied a.e. in $U$. There are many other alternative definitions. For instance, instead of working with Sobolev spaces, you can assume that $f$ is absolutely continuous on almost every coordinate line segment in $U$ (hence, has partial derivatives a.e. in $U$) and satisfies the same inequality as above. if you do not know about Sobolev spaces or absolute continuity, just think of $f$ as a diffeomorphism (this is not enough, but suffices for the intuition).

Other definitions are in terms of conformal moduli, conformal capacities, quasisymmetry,...

Some references:

Iwaniec, Tadeusz; Martin, Gaven, Geometric function theory and nonlinear analysis, Oxford Mathematical Monographs. Oxford: Oxford University Press (ISBN 0-19-850929-4/hbk). xv, 552 p. (2001). ZBL1045.30011.

Reshetnyak, Yu. G., Space mappings with bounded distortion, Translations of Mathematical Monographs, 73. Providence, RI: American Mathematical Society (AMS). xv, 362 p. (1989). ZBL0667.30018.

Väisälä, Jussi, Lectures on (n)-dimensional quasiconformal mappings, Lecture Notes in Mathematics 229. Berlin-Heidelberg-New York: Springer-Verlag. XIV, 144 p. (1971). ZBL0221.30031.

Addendum. The analogue of the Beltrami differential of a map $f$ in higher dimensions is $$ M_f(x):= J^{-2n}_f(x) (Df(x))^T Df(x), $$ a field of symmetric positive-definite matrices on $U$. The Beltrami equation $$ M_f(x)=A(x), $$ with $A(x)$ a field of positive-definite symmetric matrices on $U$, is overdetermined if $n\ge 3$ (as Alex noted in his answer). Nevertheless, this equation is sometimes useful when working with quasiconformal maps, although not as useful as the classical Beltrami equation: Most tools in higher dimensions are not analytic but geometric. An interesting thing is that if $A(x)$ is smooth, then there are known necessary and sufficient conditions for solvability of the higher-dimensional Beltrami equation (the condition is a 3rd order nonlinear PDE on $A$ if $n=3$ and a 2nd order nonlinear PDE on $A$ if $n\ge 4$). I do not know if anybody worked out a distributional analogue of this classical result.

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The significance of the Beltrami equation $f_{\overline{z}}=\mu f_z$ lies in the fact that it is solvable: for a given $\mu$, $\|\mu\|_\infty<1$ one can find $f$. For real analytic $\mu$ local solvability is essentially due to Gauss, but later it was generalized to $\mu\in L^\infty$. Moreover, it is globally solvable, in the complex plane or on the Riemann sphere, in view of the uniformization theorem. This is called the Measurable Riemann Mapping Theorem nowadays.

None of this is true in higher dimensions. One can write a similar equation but it is overdetermined and in general cannot be solved, even locally. So quasiconformal mappings in higher dimensions have to be studied with other means, an analog of the Beltrami equation plays no role there.

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