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This is a doubt of mine on the basics of complex analysis.

I encountered a certain statement involving integrating a harmonic function, which would be nice for my research attempts if proved. When I strengthened the assumption to that the function is holomorphic, I could very easily do it using Cauchy's theorem. Is it always possible to treat a harmonic function as the real or imaginary part of a holomorphic function, and draw consequences from Cauchy's theorem?

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This wikipedia article seems related. To quoting the paragraph, "The real and imaginary part of any holomorphic function yield harmonic functions on $\mathbb R^2$ (these are said to be a pair of harmonic conjugate functions). Conversely, any harmonic function $u$ on an open set $\Omega \subseteq \mathbb R^2$ is locally the real part of a holomorphic function". – Srivatsan Sep 16 '11 at 18:28
Also globally, if the open set is simply connected. – Robert Israel Sep 16 '11 at 18:46
It should be be said that you only talk about real valued harmonic functions. Linear combinations over $\mathbb{C}$ of harmonic functions are always harmonic. – AD. Sep 16 '11 at 20:47
up vote 4 down vote accepted

If $f$ is a harmonic function on a simply connected domain then it is the real or imaginary part of a holomorphic function.

If the domain is not simply connected then the above may not be true. Consider $f(x,y)=\log(\sqrt{x^2 +y^2})$ in the punctured unit disc.

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How do you find the complementary part, however? – George Sep 16 '11 at 19:04
@George, Define $h(z)$ to be the integral $-f_y dx + f_x dy$ from a fixed point to $z$. Any path is okay since the domain is simply connected, and the form is closed by harmonicity. – Soarer Sep 16 '11 at 20:07
@George locally it's the real part $\log(z)$ for any branch of the logarithm. But these local branches can't join together to form an analytic function on the whole punctured disc. – Zarrax Sep 16 '11 at 23:00
@George, I interpreted your comment as asking "If one knows the eg real part of the holomorphic function, how can we find the imaginary part?" . If so, the answer lies in the Cauchy Riemann equations. – Ragib Zaman Sep 17 '11 at 2:05
Ah, ok, thanks everyone. This was something basic and I should have always known, but didn't. Your comments helped in understanding. – George Sep 20 '11 at 7:04

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