Finding Harmonic conjugate for $\arg(z)$ 
Let $u(z)=\arg(z)$ in $D=\mathbb{C} \setminus\mathbb{R}^-$ where $\mathbb{R}^-=(-\infty,0]$. Find Harmonic conjugate for $u$.

Any ideas / hints?
 A: I think I might have a simpler solution that requires absolutely no knowledge of complex logarithms. 
If we use polar coordinates, then we get $u(r,\theta)=\theta$. Now, we want to find a function $v(r,\theta)$ so that $u(r,\theta) + iv(r,\theta)$ will be holomorphic. Meaning, we have to make sure $u$ and $v$ will satisfy the polar version of the Cauchy-Riemann equations: $$\frac{\partial u}{\partial r}=\frac 1r\frac{\partial v}{\partial \theta},\frac{\partial v}{\partial r}=-\frac 1r\frac{\partial u}{\partial \theta}$$
In our case these translate into the following: $$\frac{\partial v}{\partial \theta}=0,\frac{\partial v}{\partial r}=-\frac 1r$$
This means that $v$ is only a function of $r$ which we can solve for using simple single-real-variable integration to get: $$v(r,\theta)=-\ln r$$
A: I found it awkward to present the following argument without at least some reference to $\ln z$; so I essentially "invented" it here at equation (4), then developed a couple of basic properties such as $e^{\ln z} = z$.  But it doesn't take much "logarithm theory" to answer this question.
In $D = \Bbb C \setminus \Bbb R^- = \Bbb C \setminus (-\infty, 0]$, we may without ambiguity take
$z = re^{i\theta}, \tag 1$
since $\theta$ is restricted to the interval $(-\pi, \pi)$.  Furthermore, since the point $r = 0$ is excluded from $D$ as well, $\ln r$ is a well-defined real function on $D$.  We may therefore define the complex-valued function on $D$
$g(z) = \ln r + i\theta; \tag 2$
we note that
$e^{g(z)} = e^{\ln r + i\theta} = e^{\ln r} e^{i \theta} = re^{i\theta} = z. \tag 3$
Next, we observe that $z^{-1}$ is holomorphic in $D$, and we define the function $\ln z$ in $D$ via the formula
$\ln z = \displaystyle \int_1^z s^{-1} ds, \tag 4$
where the integral is taken over any path in $D$ which joins $1$ and $z \in D$; since $z^{-1}$ is holomorphic the integral is path-independent, and $\ln z$ is holomorphic in $D$.  We show $e^{\ln z} = z$.  Consider the holomorphic function
$F(z) = z^{-1} e^{\ln z}; \tag 5$
since
$\ln 1 = \displaystyle \int_1^1 s^{-1} ds = 0, \tag 6$
we have
$F(1) = 1; \tag 7$
also,
$F'(z) = -z^{-2}e^{\ln z} + z^{-1}(\ln z)' e^{\ln z} = -z^{-2}e^{\ln z} + z^{-2}e^{\ln z} = 0, \tag 8$
since $(\ln z)' = z^{-1}$ from (4).  I fowllows fro (8) that $F(z)$ is a constant and so by (7) we see that
$z^{-1}e^{\ln z} = F(z) = 1 \tag 9$
for all $z \in D$;  hence
$e^{\ln z} = z. \tag {10}$
We combine (2) and (10):
$e^{\ln z} = e^{g(z)}, \tag {11}$
or
$e^{(g(z) - \ln z)} = 1; \tag {12}$
if follows that
$g(z) = \ln z + 2n\pi i \tag{13}$
for some $n \in \Bbb Z$; but
$g(1) = 0 = \ln 1, \tag{14}$
so $n = 0$ and
$\ln r + i \theta = g(z) = \ln z \tag{15}$
in $D$.  (15) shows that $\ln r + i \theta$ is holomorphic in $D$ and thus
the harmonic conjugate of $\theta = \arg(z)$ in $D$ is $\ln r$.
A: Let the function be $$\arctan\frac yx+i v(x,y).$$
The Cauchy-Riemann equations read
$$v'_y(x,y)=-\frac y{x^2+y^2},\\
v'_x(x,y)=-\frac x{x^2+y^2}.$$
If we integrate the first on $y$,
$$v(x,y)=-\log\sqrt{x^2+y^2}+c(x).$$
We now plug it in the second,
$$-\frac x{\sqrt{x^2+y^2}}+c'(x)=-\frac x{x^2+y^2}.$$ This shows that $c(x)$ must be a constant.
Finally,
$$f(z)=\arctan\frac yx-i\log\sqrt{x^2+y^2}+ic,$$ which you can recognize to be $$-i\log z+ic.$$
