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?
The function $f$ you have obtained is nothing but $f(z) =\frac 1 z +iC$ which is holomorphic.