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Suppose $f:\mathbb{C}\to\mathbb{C}$ is an analytic function and $f:=u+iv$. Then is it always true that $u_{xx}+u_{yy}=v_{xx}+v_{yy}=0$ ?

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    You know that analytic functions satisfy the Cauchy-Riemann equations, right? Take those and try to proceed from there.2012-08-20
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    But for that we need $v_{xy}$ and $v_{yx}$ to be continous, is that always true ?2012-08-20
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    An analytic function is a power series, and thus its derivatives are all smooth.2012-08-20
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    I got the answer. In wikipedia I saw that analytic functions are infinitely differentiable. Thanks for all your comments.2012-08-20
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    @pritam, we need to assume $u_{xy}=u_{yx}$ and $v_{xy}=v_{yx}$, right? I'd like to know in what condition $f_{xy}≠f_{yx}$ and when they coincide.2012-08-20
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    @lab $f_{xy}=f_{yx}$ when they are both continous. See here :http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Clairaut.27s_theorem2012-08-20

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