# is the converse true: in a simply connected domain every harmonic function has its conjugate

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.

-
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

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.

-
:-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

-
Welcome to math.stackexchange.com! Great first post! –  Hans Engler May 9 '14 at 18:31