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I understand that Maass forms are the eigenfunctions of the Laplacian that are smooth on $\mathbb{H}$ but not holomorphic. But I can't write down an explicit formula for such a function. How might I reconcile the intuition that this shouldn't be possible since all smooth functions on the complex plane are holomorphic?

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Any complex-differentiable function is holomorphic, but smooth functions are only required to be (infinitely) real-differentiable (i.e., its real and imaginary parts are both (infinitely) differentiable in the sense of multivariable calculus as functions $\mathbb{R}^2\to\mathbb{R}$). A very simple example of a smooth function that is not holomorphic is $\operatorname{Re} z$.

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  • $\begingroup$ Oh oops. Thanks. $\endgroup$ – Tommy Tang Oct 30 '15 at 5:23

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