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I have read a theorem that states

If $\Omega$ is a simply connected domain and $u:\Omega\rightarrow\mathbb{R}$ is harmonic, then there exists a harmonic function $v:\Omega\rightarrow\mathbb{R}$ such that $f$, given by $f(x+iy)=u(x,y)+iv(x,y)$ in $\Omega$, is holomorphic.

My question is, what if the domain is not simply connected? How can we determine the harmonic conjugate of $v$ (if one exists)? For example, consider $$u(x,y)=\frac{x}{x^2+y^2}.$$ Now $u(x,y)$ is harmonic in $\mathbb{C}$\ $\{0,0\}$. How can i determine its harmonic conjugate? To find $v(x,y)$, I took the assumption that $f(x+iy)=u(x,y)+iv(x,y)$ was holomorphic and calculated that $$v(x,y)=-\frac{y}{x^2+y^2}+C.$$But how can I be certain that this is correct and that $f$ is in fact holomorphic in the first place? Keeping in mind I am yet to study complex integration (path integrals, etc).

Another example is the function $$u(x,y)=\frac{1}{2}\ln|x^2+y^2|$$ Now I know that the harmonic conjugate of $v(x,y)=tan^{-1}\left(\frac{y}{x}\right)$ should be $u$, as this is the imaginary part of the complex logarithm. But assuming I did not know this, how could I determine if $v(x,y)$ is harmonic (does it have to exist?) and hence a function $f(x,y)=u(x,y)+iv(x,y)$ which is holomorphic?

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    $\begingroup$ Possible duplicate of Link between harmonic and holomorphic functions on a non-simply connected domain. $\endgroup$ – mfl Aug 16 '18 at 11:43
  • $\begingroup$ Thanks for that, I did not see the previous post. However, the response in this question seems to rely on the use of complex integration (e.g. Morera's Theorem), which I have not studied this yet. Is there another way to explain this? $\endgroup$ – user557493 Aug 16 '18 at 11:56
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    $\begingroup$ Actually it is not a duplicate. I have misread the question and thus I voted duplicate. I have retracted my vote. I am sorry. $\endgroup$ – mfl Aug 16 '18 at 12:03
  • $\begingroup$ I shall include another example, which may possibly help. $\endgroup$ – user557493 Aug 16 '18 at 12:04
  • $\begingroup$ But it is a possible duplicate of this one math.stackexchange.com/questions/108116/… If you have a look of the answer you'll get your answer. $\endgroup$ – mfl Aug 16 '18 at 12:12
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The function $f$ you have obtained is nothing but $f(z) =\frac 1 z +iC$ which is holomorphic.

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    $\begingroup$ Yes, I am aware of that. However that was not my question. My question is how do I know $f$ is holomorphic given the domain of $u$ is not simply-connected. If I start with the (incorrect) assumption that the domain of $u$ is simply-connected and then proceed with following calculations, I arrive at $f(z)=\frac{1}{z}$ just as you did. I guess another way of asking this question is how do I find a harmonic conjugate of $u$, given the domain of $u$ is not simply-connected. $\endgroup$ – user557493 Aug 16 '18 at 12:02
  • $\begingroup$ In a simply connected region every harmonic function has a harmonic conjugate but in other regions there may be no harmonic conjugate. But if a harmonic function has a harmonic conjugate in a non-simply connected region you compute the conjugate the same way you do in a simply connected region. The conjugate does give you a holomorphic function and this function does satisfy C-R equations. What makes you think computation of the conjugate has to be more complicated (or different) in non-simply connected regions? $\endgroup$ – Kavi Rama Murthy Aug 16 '18 at 12:38
  • $\begingroup$ Thank you for that piece of information. I just don't understand the distinction between finding a function $v$ such that $v$ is the harmonic conjugate of $u$ in a domain that is simply-connect and not simply-connected. So if the process is the same, how do I know if a harmonic function $v$ exists? For instance, for the example above, how do I know that a $v$ exists such that $v$ is the harmonic conjugate of $u(x,y)=\frac{x}{x^2+y^2}$ and consequently $f$ is holomorphic? $\endgroup$ – user557493 Aug 16 '18 at 12:54
  • $\begingroup$ There is no thumb rule to determine if a given harmonic function has a harmonic conjugate. You have try solving the C-R equations. $\endgroup$ – Kavi Rama Murthy Aug 16 '18 at 23:48
  • $\begingroup$ This was very helpful :) I similar question came up in my exam and I was able to solve it, thanks to your help. $\endgroup$ – user557493 Sep 3 '18 at 6:37

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