$u$ harmonic can be represented as sum of $f+\bar{g}$, $f,g$ holomorphic. On this pdf (page 5): 

If $u$ is harmonic on a disk then it can be represented as $f+\bar{g}$
  where $f,g$ are holomorphic.

I could not follow this line of proof:
What does mean to integrate wirtinger derivatives, and wrt $\bar{z}$? May someone elaborate the details?
 A: Working backwards using the definitions for Wirtinger derivatives and complex differentials 
\begin{align}
&\int_{\gamma} \frac{\partial}{\partial z} u(z) dz+\int_{\gamma} \frac{\partial}{\partial \bar z} u(z) d\bar z &= \\
&= \int _{\gamma} \frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{1}{i}\frac{\partial}{\partial y}\right) u(x,y) (dx+idy) + \int _{\gamma} \frac{1}{2}\left(\frac{\partial}{\partial x}-\frac{1}{i}\frac{\partial}{\partial y}\right) u(x,y) (dx-idy)\\
&= \int_{\gamma}\frac{1}{2}\frac{\partial}{\partial x} u(x,y)dx +\int_{\gamma}\frac{1}{2i}\frac{\partial}{\partial y} u(x,y)dx +\int_{\gamma}\frac{i}{2}\frac{\partial}{\partial x} u(x,y)dy +\int_{\gamma}\frac{1}{2}\frac{\partial}{\partial y} u(x,y)dy \\ 
&+ \int_{\gamma}\frac{1}{2}\frac{\partial}{\partial x} u(x,y)dx
-\int_{\gamma}\frac{1}{i}\frac{\partial}{\partial y} u(x,y)dx  
-\int_{\gamma}\frac{i}{2}\frac{\partial}{\partial x} u(x,y)dy 
+\int_{\gamma}\frac{1}{2}\frac{\partial}{\partial y} u(x,y)dy \\ 
&=\int_{\gamma} \frac{\partial}{\partial x} u(x,y) dx+\int_{\gamma} \frac{\partial}{\partial y} u(x,y) dy \\
&= \int_0^{1}u'(\gamma(t))\gamma '(t)dt \\
&= \int_0^{1}\frac{d}{dt}u(\gamma(t))dt
\end{align}
