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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. – timur Aug 24 '12 at 20:36
Let $u:\Omega\subseteq \mathbb{C}\rightarrow \mathbb{R}$ be harmonic and has conjugate, does it imply $\Omega$ is simply connected? – miosaki Aug 24 '12 at 20:38
$\Omega$ is simply connected if every real harmonic function on $\Omega$ has a conjugate. – Jonas Meyer Aug 24 '12 at 20:49
up vote 3 down vote accepted

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 :-o – miosaki Aug 24 '12 at 20:41
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. – JSchlather Aug 24 '12 at 21:22
@JacobSchlather I couldn't get :-o – miosaki Aug 24 '12 at 21:49

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! Great first post! – Hans Engler May 9 '14 at 18:31

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