General way to express holomorphic function in terms of z? For the holomorphic, complex-valued function f, defined as
$f(x + iy) = xy - x + y + i(-(1/2)x^2 + (1/2)y^2 - x - y + c)$
We can express this in terms of $z$ and $\bar z$  by substituting
$x = (1/2)(z + \bar z)$
and
$y = -i(1/2)(z - \bar z)$,
which puts f in the form
$f(z) = -i(1/2)z^2 - (1 + i)z + ic$, where we can see it does not depend on $\bar z$.
How is this substitution in particular chosen, and how would a valid/useful substitution be chosen for an arbitrary f? 
Is there more than one possible substitution that could be performed in either this or the general case?
 A: If you already know that $f(x+iy)$ is holomorphic, which you can check using Cauchy-Riemann's equations, there is a nice, efficient trick to express $f$ in terms of $z$ (no $\bar z$ needed, the function is holomorphic!):

Put $y=0$ and $x=z$. This gives you $f(z)$.

For your particular example:
\begin{align}
f(z+i0) &= z\cdot 0 - z + 0 + i(-(1/2)z^2 + (1/2)0^2 - z - 0 + c) \\
&= \frac{-i}{2} z^2 - (1+i)z + ic.
\end{align}
Why does this work?
Let $g(z)$ be the function constructed by the trick above. Then $g$ is holomorphic, and $g(x) = f(x)$ whenever $x$ is real (by construction). The identity theorem for holomorphic functions shows that $f = g$ everywhere.
A: Functions in terms of the real or imaginary parts of a number are a broader set than those on the number itself. So as you might have guessed, there is no way to always eliminate the $\overline{z}$ from the equation. One example that shows this is f(x+iy)=x. 
However, if having $\overline{z}$ is alright, you can always convert from x and y to z and $\overline{z}$. This is simply a reparameterization (you can think back to polar coordinates, when you converted from x and y to r and $\theta$)
