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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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1  
You know that analytic functions satisfy the Cauchy-Riemann equations, right? Take those and try to proceed from there. –  t.b. Aug 20 '12 at 7:49
    
But for that we need $v_{xy}$ and $v_{yx}$ to be continous, is that always true ? –  pritam Aug 20 '12 at 8:01
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An analytic function is a power series, and thus its derivatives are all smooth. –  Siminore Aug 20 '12 at 8:07
    
I got the answer. In wikipedia I saw that analytic functions are infinitely differentiable. Thanks for all your comments. –  pritam Aug 20 '12 at 8:35
    
@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. –  lab bhattacharjee Aug 20 '12 at 8:39

1 Answer 1

Analytic function satisfies Cauchy Riemann equation.

i.e.

$u_{x}=v_{y}$ & $u_{y}=-v_{x}$

so, $u_{xx}=v_{xy}$ & $u_{yy}=-v_{yx}$

so, $u_{xx}+u_{yy}=0\ when\ v_{xy}=v_{yx}$

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