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.
Best 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)