# Show that $\log \left| z \right|$ is harmonic and find its the conjugate harmonic function.

Is the form correct for the conjugate harmonic?

Attempt:

First, we are given \begin{align*} \log \left| z \right| &= u(x,y) + iv(x,y) = \log \sqrt{x^2 + y^2} + i \cdot 0 \\ u(x,y) &= \log \sqrt{x^2 + y^2} = \frac{1}{2} \log (x^2 + y^2). \end{align*} Then we differential to get $u_{xx}$ and $u_{yy}$, \begin{align*} u_{xx} = \frac{\partial u}{\partial x} \frac{x}{x^2 + y^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} \\ u_{yy} = \frac{\partial u}{\partial y} \frac{y}{x^2 + y^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2} \end{align*} From here, we can see that $u_{xx} + u_{yy} = 0$. Thus, we have shown that $\log \left| z \right|$ is harmonic. Using the Cauchy-Riemann equations, we can find the conjugate harmonic function $v$. This gives us the relationships \begin{align*} u_x &= v_y = \frac{x}{x^2 + y^2} \\ u_y &= -v_x = \frac{y}{x^2 + y^2}. \end{align*} Integrate with respect to $y$ to get $v(x,y) = \tan^{-1}(\frac{y}{x}) + C$, the conjugate harmonic function.

## 1 Answer

Almost correct. $log|z|$ have a harmonic conjugate iff $z$ is in some simply connected subset of $\mathbb C \backslash \{ 0 \}$ (think of convex subsets). I think you $v$ should be like the $\text{Arg}$ function in http://en.wikipedia.org/wiki/Argument_(complex_analysis)