Find the harmonic conjugate of $u$. $u = u(z) = \ln(|z|)$ so $u(z) = \ln(\sqrt{x^2 + y^2})$
I am trying to find now its harmonic conjugate I did all the math:
I got two solutions though. One is $v(z) = \arctan(y/x) + C$ if I solve $\partial u/\partial x = -\partial v/\partial y$ & other is $v(z) = - \arctan(x/y) + C$ if I solved $\partial v/\partial y = \partial u/\partial x$
can I have two solutions, or is only one solution correct?
$\arctan(y/x) + \arctan(x/y) = \pm \frac\pi2 $. So in essence, your two solutions are the same.
Another way to get your solution: $\ln(z)$ is analytic, and if $z = r e^{i\theta}$, then $\ln(r e^{i\theta}) = \ln r + i\theta$. And $\theta$ is the angle $z$ makes to the $x$-axis, that is, $\arctan(y/x)$. You can get the other solutions by considering, for example, $\ln(iz)$, or more generally $\ln(az)$ where $|a| = 1$.
A harmonic function is a twice differentiable function that satisfies Laplace's equation. Recall that for a function to be differentiable, it must first be continuous.
As you found out by going through the calculations, the harmonic conjugate $v(z)$, if it exists, must be either $v(z) = \arctan(y/x) + C_1(x)$ (where $C_1$ is a constant with respect to $y$ but is possibly a non-constant function of $x$) or $v(z) = \arctan(x/y) + C_2(y)$.
However, neither choice is continuous on $\mathbb{R}^2$ or $\mathbb{C}$, so neither choice is harmonic. Therefore $u(z)$ does not have a harmonic conjugate.
To see that neither choice is continuous, consider the limit $(x, y) \to (0, 0)$ of $\arctan(y/x)$ and $\arctan(x/y)$ along the lines $y = x$ and $y = -x$. Along the first line, they are both equal to $\arctan(1) = \sqrt{3}$; along the second line they are both equal to $\arctan(-1) = -\sqrt{3}$.
