# harmonic conjugate of a natural log

Find the harmonic conjugate of $\ln \sqrt{x^2+y^2}$ on some open nonempty subset of the plane.

Ok I got stuck. So I set the function as $u(x,y)=\ln \sqrt{x^2+y^2}$. So to show that $u_x(x,y)= v_y(x,u)$ I found the derivative of $u_x= \frac{x}{x^2+y^2}$, so that I can write $v_y= \frac{x}{x^2+y^2}$.

Ok so now I'm trying to find $v(x,y)$ but I seem to be getting stuck on the math part here. I know I have to get the integral to get it in terms of y but it doesn't seem to be coming out for me. Integration by parts didn't pan out for me. Please help I would really appreciate it.

• $$\frac{x}{x^2+y^2} = \frac{x}{x^2\left(1+\left(\frac{y}{x}\right)^2\right)}= \frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \frac{1}{x}$$ Commented Mar 14, 2014 at 6:50
• that answer looks a bit odd. Shouldn't it look somewhat like the original function in a way Commented Mar 14, 2014 at 7:26

We can use the identity $$\tan(\theta/2)=\frac{\sin(\theta)}{1+\cos(\theta)}$$ to extend the arctan into quadrants II and III: $$\log(x+iy)=\log\left(\sqrt{x^2+y^2}\right)+2i\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right)$$ This only fails when $x+\sqrt{x^2+y^2}=0$, which is only when $y=0$ and $x\le0$. This corresponds to a branch cut for $\log(z)$.