Harmonic function as the sum of two holomporphic functions Let $u:\mathbb{C} \rightarrow \mathbb{R}$.
Prove that $u$ is an harmonic function iff $u(z) = f(z) + g(\bar{z})$ where $f,g$ are holomorphic functions.
I'd be happy for a hint
Thanks!
 A: The "if" part follows from Cauchy-Riemann equations immediately. To show the "only if" part, by considering the Cauchy-Riemann equations, please try to find the conjugate harmonic function $v$ of $u$, which satisfies that $h=u+iv$ is holomorphic.
Remark: There are several ways to construct $v$(or $h$ equivalently)  once $u$ is given. One way is to define $v$ as a line integral: $v(z)=\int_0^z(u_ydx-u_xdy)$, where the value of the integral is independent of the choice of path, because $u$ is harmonic on $\mathbb{C}$. An alternative way is based on the following observation. If $h=u+iv$ is holomorphic, then $h'=u_x+iv_x=u_x-iu_y$. Now given $u$ harmonic on $\mathbb{C}$, $u_x-iu_y$ satisfies Cauchy-Riemann equations, so it must be holomorphic. Then there is a unique $h:\mathbb{C}\to\mathbb{C}$, holomorphic, such that $h'=u_x-iu_y$ and $h(0)=u(0)$. Note that $u$ is the real part of $h$, so $f$ and $g$ can be constructed in terms of $h$. 
A: There is another explanation of why this should be true.
First consider the corresponding "real-valued" problem. If you have a smooth function of two variables, $U:\mathbb{R}^2 \to \mathbb{R}$, such that $\frac{\partial^2}{\partial x \partial y} U(x,y) = 0$, it is easy to show that $U(x,y) = F(x) + G(y)$ for some $F,G:\mathbb{R} \to \mathbb{R}$ by constructing such functions explicitly.
Now, since $\Delta = 4 \frac{\partial^2}{\partial z \partial \overline z} $, and $z$ and $\overline z$ are algebraically independent, you may try to do the same with a harmonic function. The only technical thing you need is that any harmonic function is real analytic, hence it has an analytic continuation to a neighborhood of $\mathbb{R}^2 \subset \mathbb{C}^2$. Now you should just express this analytic function $U$ of $x$ and $y$ in terms of $z=x+iy$ and $z^\prime=x-iy$, essentially treating $z$ and $z^\prime$ as independent (of course, in $\mathbb{R}^2$ we have $z^\prime = \overline{z}$, but this no longer holds in $\mathbb{C}^2$). Observe that $\frac{\partial^2}{\partial z \partial z^\prime} U = 0$, and try to do the same thing that you would do in the real case, which should yield $U(z,z^\prime) = F(z)+G(z^\prime)$ with $F$ and $G$ complex analytic. Then restrict them back to $\mathbb{R}^2$.
