Find the holomorphic function $f(z)=u+iv$ the most general 
Show that $v(x,y)=xy$ is a harmonic function (done) and find the
  holomorphic function $f(z)=u+iv$ the most general.

We know that if $f$ is holomorphic, than it must satisfy the Cauchy-Riemann equations. So $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=x$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}=-y$, and I found the function $f(z)=(\frac{x^2}{2}+c)+ixy$. Am I right or I have to modify something?
 A: If $u_x(x,y) = x$, then $u(x,y) = \frac{x^2}{2}+h(y)$. Then $u_y(x,y) = h'(y) = -y$, whence $h(y) = -\frac{y^2}{2}+c$. So $u(x,y) = \frac{x^2-y^2}{2}+c$ and $$f(x,y) = \frac{x^2-y^2}{2}+c + ixy.$$The mistake was the following: when you integrated $u_x(x,y) = x$, the "constant" of integration is not actually a constant, but a function of the other variables. 
Finally, note that for $c = 0$, we've found $f(z) = z^2/2$.
A: This question has been considered from the point of view of differential equations issued from Cauchy-Riemann conditions.
I would like to point out a geometrical aspect of the question.
First of all, I advise you to have a look at Fig. 7 (left) of the interesting lecture notes http://www.math.umn.edu/~olver/ln_/cml.pdf
The curves in red are $xy=$const., the curves in blue are $x^2-y^2=$ const., i.e., the level curves of your $u$ and $v$ (hyperbolas in both cases). These level curves are othogonal: this is a completely general property. 
If you go further in the lecture notes of Olver, you fill see how all this can be "seen" for example in terms of fluid dynamics, displying a kind of "duality" between the two conjugate functions $u$ and $v$.
