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Question: If harmonic functions $u$ and $v$ satisfy Cauchy-Riemann equations, then $u+iv$ is an analytic function.

Am a bit confused here as we already have a theorem which says that if a function satisfies Cauchy Riemann equations for each point in the domain and each of its four partial derivatives $u_x, v_x, u_y, v_y$ are continuous then it is analytic in that domain.

So can we say that if $u$ and $v$ are harmonic then each of its four partial derivatives $u_x, v_x, u_y, v_y$ are continuous thereby implying that the function is analytic in $D.$

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There are harmonic functions that do not satisfy the Cauchy-reimann equations. (x+y)+i(x-y) is an example.

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Harmonic functions $u:\mathbb{R}^n\rightarrow \mathbb{R}$ are known to be smooth (i.e. infinitely often continuously differentiable). This is, however, not a completely trivial statement (and the definition of harmonic may be weakened considerably from what is probably known to you).

In most cases, in two dimension, it should be sufficient to know that if a function $u$ is continuously differentiable and you can 'write down $\Delta u$' then the conclusion of the theorem you cited holds true. This is particularly interesting if you have a harmonic function $u$ and want to know whether it's smooth (or even real analytic). The theorem you are using tells you that you can conclude this already if you find a complex analytic extension $u+iv$ which is complex analytic.

Note however that $u, v$ both being harmonic does not imply that $u+iv$ is complex analytic. The $v$ you need for given $u$ is a special one (and locally uniquely determined up to a constant).

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  • $\begingroup$ See 5th line of en.wikipedia.org/wiki/Harmonic_function#Remarks Here it says that all harmonic functions are analytic. So, if both $u$ and $v$ are harmonic (implies analytic), then why isn't $u + iv$ analytic? $\endgroup$ – user265328 Dec 8 '15 at 16:42
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    $\begingroup$ @santa there is a difference between the concepts of being real analytic or complex analytic. A function is real analytic if can be represented everywhere by a (real valued) powerseries. This is a property which is usually extremly hard to verify, the case of harmonic functions is a special case where it is easy to see. A function is complex analytic if is holomorphic, i.e. complex differentiable. This implies, too, it can be written as a (complex and complex valued) power series , but this is a stronger property than requesting that both complex and imaginary part are real analytic $\endgroup$ – Thomas Dec 8 '15 at 17:43
  • $\begingroup$ Interesting.. Thanks for the clarification. $\endgroup$ – user265328 Dec 9 '15 at 13:23

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