Finding Conjugate harmonic of $u = \frac{1}{2} ln(x^2 + y^2)$ this is a nice community;
I've been facing a hard time answering this question, a detailed answer would be splendid.
$u = \frac{1}{2} ln(x^2 + y^2)$
find conjugate harmonic, and harmonic function.
Thanks very much
Edit: this is my solution so far
found ∂u∂x, it's square, and ∂u∂y with its square
tried using this equation (F(z)= integration(∂u∂x(z,0)- j∂u∂y(z,0))dz + jc)
....but couldn't get a sensible result of (v)
my results was F(z)=ln(z) +jc = ln(x+jy) + jc ..... as you can see this is not a satisfying solution
 A: Let $u(x,y)=\frac12 \log(x^2+y^2)$ for $(x,y)\ne(0,0)$.  If $v(x,y)$ is the harmonic conjugate of $u(x,y)$, then there is an analytic function $f(z)$ such that $f(z)=u(x,y)+iv(x,y)$.  
If $f(z)$ is analytic, then $u(x,y)$ and $v(x,y)$ satisfy the Cauchy-Riemann Equations
$$\frac{\partial u(x,y)}{\partial x}=\frac{\partial v(x,y)}{\partial y} \tag 1$$
and
$$\frac{\partial u(x,y)}{\partial y}=-\frac{\partial v(x,y)}{\partial x} \tag 2$$
From $(1)$, we find that
$$\frac{\partial v(x,y)}{\partial y}=\frac{x}{x^2+y^2}\tag 3$$
Integrating $(3)$ with respect to $y$, we obtain
$$v(x,y)=\arctan2(y,x)+C(x) \tag 4$$
where $C(x)$ is an integration constant with respect to $x$ and $\arctan2(y/x)$ is defined as in THIS ARTICLE.
Applying the Cauchy-Riemann Equation $(2)$ to $(4)$ reveals
$$\frac{\partial u(x,y)}{\partial y}=\frac{y}{x^2+y^2}+C'(x)$$
from which we find $C'(x)=0$.  
Therefore, setting the integration constant to $0$, the harmonic conjugate of $u(x,y)=\frac12 \log(x^2+y^2)$ is 
$$v(x,y)=\arctan2(y,x)$$
Putting it all together, 
$$f(z)=\frac12 \log(x^2+y^2)+i\arctan2(y,x)=\log(z)$$
