Poisson equation using complex analytic method I have the following question in a complex analysis text: 
Find a particular solution to the following Poisson equation: 
$$\nabla^2u(r, \theta) = r^2 \cos \theta.$$ 
The solution method outlined in the text uses Wirtinger derivatives to simplify the equation, then integrate twice. So here we would have: 
$$4 \frac{\partial^2u}{\partial z \partial \bar{z}} = z \bar{z} \cos (\arg z).$$ 
Integrating with respect to $\bar{z}$ would then give: 
$$4 \frac{\partial u}{\partial z} = z \left( \frac{\bar{z}^2}{2} \right) \cos (\arg z) + f(z).$$ 
Now, here I'm stuck. I can't get this integration to work, in part because I'm not sure you can integrate $\arg(z)$. Any pointers on where I'm going wrong would be greatly appreciated. Thanks in advance. 
 A: If $u$ is a solution to this one and $v$ is a solution to $\nabla^2 v(r,\theta) = r^2 \sin \theta)$, then $w = u + i v$ satisfies
$$ 4 \dfrac{\partial^2 w}{\partial z \partial \overline{z}} = \nabla^2 w = r^2 \exp(i\theta) = r z= z^{3/2} \overline{z}^{1/2}$$
Integrate with respect to $z$ and $\overline{z}$, and you find one solution is
$$ w  = \frac{z^{5/2} \overline{z}^{3/2}}{15} = \frac{r^{3} z}{15} = \frac{r^4}{15} \exp(i\theta)$$
So taking the real part, $$u = \frac{r^4}{15} \cos(\theta)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\cos\pars{\theta} = {x \over r} =
{z + \bar{z} \over 2\root{\vphantom{\large a^{a}}z\bar{z}}}\quad}$ and
  $\ds{\quad\mrm{u}\pars{r,\theta} =
\mrm{U}\pars{r\expo{\ic\theta},r\expo{-\ic\theta}}}$.

\begin{align}
4\,{\partial^{2}\mrm{U}\pars{z,\bar{z}} \over \partial z\,\partial\bar{z}} & =
z\bar{z}\,{z + \bar{z} \over 2\root{\vphantom{\large a^{a}}z\bar{z}}} =
{1 \over 2}\pars{z^{3/2}\,\bar{z}^{1/2} + z^{1/2}\,\bar{z}^{3/2}}
\\[5mm]
\partiald{\mrm{U}\pars{z,\bar{z}}}{z} & =
{1 \over 8}\pars{z^{3/2}\,{2 \over 3}\,\bar{z}^{3/2} +
z^{1/2}\,{2 \over 5}\,\bar{z}^{5/2}} + \mrm{f}\pars{z}
\\
\mrm{U}\pars{z,\bar{z}} & =
{1 \over 8}\pars{{2 \over 5}\,z^{5/2}\,{2 \over 3}\,\bar{z}^{3/2} +
{2 \over 3}\,z^{3/2}\,{2 \over 5}\,\bar{z}^{5/2}}\ +\
\overbrace{\int\mrm{f}\pars{z}\dd z}^{\ds{\equiv\ \mrm{F}\pars{z}}}
\\[5mm]
\mrm{U}\pars{z,\bar{z}} & =
{1 \over 15}\,\Re\pars{z^{5/2}\,\bar{z}^{3/2}} + \mrm{F}\pars{z} =
{1 \over 15}\,\pars{z\,\bar{z}}^{3/2}\,\Re\pars{z} + \mrm{F}\pars{z}
\\[2mm] & =
\bbx{{1 \over 15}\,r^{4}\cos\pars{\theta} + \mrm{F}\pars{r\expo{\ic\theta}}}
\end{align}
