Solving cauchy riemann equations, finding all analytic functions I need someone to check my work! I tried doing this as properly as possible, but I have no way to check whether this is correct.
Find $\textit{all}$ analytic functions $f = p(x,y) +iq(x,y) $ such that $p+q = xy$
(I'm using p's and q's since u's and v's look too similar. It gets messy quickly).
We have $$p_x + q_x = y \qquad p_y + q_y = x $$
$f$ is to be analytic so by Cauchy-Reimann we have $p_x = q_y$ and $p_y = -q_x$, thus we have:
$$y- q_x = x - p_y \qquad y+p_y = x- p_y$$
so 
$$ p_y = \frac{x-y}2$$
Thus $$p = \frac{xy}{2} - \frac{y^2}{4} + h(x)$$
Since $q = xy - p$ we have 
$$ q = xy - \frac{xy}{2} + \frac{y^2}{4} - h(x) = \frac{xy}{2}+ \frac{y^2}{4} - h(x) $$
Again, since $p_x = q_y$ we have
$$\frac{y}2 + h'(x) = x - \frac{x}2 + \frac{y}2 = \frac{x}2 + \frac{y}2$$
so we have: 
$$h(x) = \frac{x^2}4 + C $$
Putting it all together: 
$$ f = p+iq = \frac{xy}{2} - \frac{y^2}{4} + \frac{x^2}4 + C + i\left( \frac{xy}{2}+ \frac{y^2}{4} - \frac{x^2}4 - C \right)$$
If we would like to rewrite this, we have:
$$f(z) = \frac{z^2}4 + C - i\left( \frac{z^2}4 + C \right) = (1-i)\left( \frac{z^2}4 + C \right)  $$ 
Now I would like to claim that for $C \in \mathbb{R}$, these are all solutions. Can I truly be sure of this? Did I make some mistake somewhere?
Thanks in advance.
 A: It is of course possible that I overlooked a mistake, but since the result is correct, and I didn't see any error, I'm reasonably convinced that your work is correct.
Here, as in many situations, we can solve the task in an easier way if we apply a certain transformation to the problem. It is advisable to try to look for such simplifications. Not only does it save work when one spots them, it is also less likely to make mistakes in simpler situations.
With $f = p + iq$, note that $(1+i)f = (p-q) + i(p+q)$, so $xy$ is the imaginary part of $g = (1+i)f = u + iv$. Now the Cauchy-Riemann equations make it easy to find $u$, since
$$u_x = v_y = x \quad \text{and}\quad u_y = -v_x = -y,$$
so $u(x,y) = \frac{1}{2}(x^2 - y^2) + c$ for some $c\in \mathbb{R}$. Writing $g$ as a function of $z = x+iy$, we find
$$g(z) = \frac{z^2}{2} + c$$
for some (arbitrary) $c\in \mathbb{R}$. Since $(1-i)(1+i) = 2$, we then obtain
$$f(z) = \frac{1-i}{2} g(z) = \frac{1-i}{4} z^2 + \frac{1-i}{2}c,$$
with an arbitrary real $c$. (My $\frac{c}{2}$ corresponds to your $C$.)
