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Let $U \subset \mathbf C$ be an open subset of the complex plane and suppose we have a differential operator of order 1, $L: \mathcal C^{\infty}(U) \to C^{\infty}(U)$ such that $Lu = 0$ if and only if $u$ is holomorphic in $U$. Is it true that $L$ must be the Cauchy-Riemann operator $\frac{\partial}{\partial \bar z}$?

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    @Lucas: If by *first order differential* you mean something of the form $au_x + bu_y + cv_x + cv_y$, $a,b,c,d\in C^\infty(\mathbb{C})$, and $u, v\in C^\infty(\mathbb{C})$ such that $f = u + iv$, then the answer is of course *no*, since $u_x - v_y$ is$a$counterexample. If you meant in the sense of what Davide suggested, then the answer is *yes* up to multiplication by a smooth function: take $f(z) = z$ and apply $a(z)\partial_z + b(z)\partial_{\overline{z}}$ to it; you'll get $a(z) = 0$, hence the original operator is $b(z)\partial_{\overline{z}}$. If you meant something else, please clarify.2012-05-18

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