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The question is.

Is the converse true: In a simply connected domain every harmonic function has its conjugate?

I am not able to get an example to disprove the statement.

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    What is the converse? Please state it explicitly.2012-08-24
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    Let $u:\Omega\subseteq \mathbb{C}\rightarrow \mathbb{R}$ be harmonic and has conjugate, does it imply $\Omega$ is simply connected?2012-08-24
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    $\Omega$ is simply connected if every real harmonic function on $\Omega$ has a conjugate.2012-08-24

2 Answers 2

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No, to the question as stated by OP in the comments. Take the real part of a holomorphic function on an annulus.

On the other hand, look at the comment by Jonas.

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    :-o :-o :-o :-o :-o :-o2012-08-24
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    Do you have any idea about the other converse? If every harmonic function on a domain $\Sigma$ has a conjugate then $\Sigma$ is simply connected.2012-08-24
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    @JacobSchlather I couldn't get :-o2012-08-24
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The answer is yes. If a domain is not simply connected, you can always construct a harmonic function without a harmonic conjugate there. For example, for an annulus centred at the origin, take $f(x,y)=\log\sqrt{x^2+y^2}$ in . This cannot have a harmonic conjugate, as if it did you would get a branch of the logarithm analytic in such a domain.

You can look at a nice explanation for the construction in the general case here:

Show $\Omega$ is simply connected if every harmonic function has a conjugate

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    Welcome to math.stackexchange.com! Great first post!2014-05-09