Going from $f = u(x,y) +iv(x,y)$ to $f(z) = f(x+iy)$ Quick question: 
Quite often, when doing stuff in Complex Analysis, I'm asked to put something of the form $f = u+iv$ into the form $f(z)$. I HATE this step, because it always amounts to me just looking at it, and trying to sorta guess half way, and work backwards from the guess. It's time-consuming, tedious and clumsy. This will not do.
I'm now looking for appropriate theory (in the ideal case, an algorithm), or some other simple approach that will remove much of the guess work, remove stuff like the 'magically noticing' obscure trig identities, and just make the whole thing a bit more palatable.
What are some good tactics? 
Thanks in advance.
 A: Firstly I always check if it satisfies Cauchy-Riemann equations or not.
If it satisfies you can write $f(x,y) = u(x,y)+iv(x,y)$ into the form $f(z)$. If it does not satisfy Cauchy-Riemann equations, you cannot write the $f(x,y) = u(x,y)+iv(x,y)$  into the form $f(z)$.
I would like to show my strategy in some examples

Example 1:  ( you cannot convert the $f(x,y)$ into $f(z)$ in this example because it does not satisfy Cauchy-Riemann equations. Thus  you do not need to struggle for converting into $f(z)$  because you cannot in any way)
$$f(x,y)=x^2+y^2+i2xy$$
$$u(x,y)=x^2+y^2$$
$$v(x,y)=2xy$$
Cauchy-Riemann equations:
$$\frac{\partial{u}}{\partial{x}}=\frac{\partial{v}}{\partial{y}}$$
$$\frac{\partial{u}}{\partial{y}}=-\frac{\partial{v}}{\partial{x}}$$
$$2x=2x$$
$$2y \neq -2y$$  Thus you cannot do transform for example 1.

Example 2:  (The example you can convert the $f(x,y)$ as $f(z)$)
$$f(x,y)=\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}$$
$$u(x,y)=\frac{x}{x^2+y^2}$$
$$v(x,y)=-\frac{y}{x^2+y^2}$$
If you check, It satisfies Cauchy-Riemann equations thus you can convert into  form of $f(z)$ 
Thus use the known relation for  $z$ for convertion .
We know $z=x+iy$ so
 $x=z-iy$ 
Put it into the equation and you will see that $y$ will disappear after operations because it satisfies Cauchy-Riemann equations 
$$f(z)=\frac{z-iy}{(z-iy)^2+y^2}-\frac{iy}{(z-iy)^2+y^2}$$
$$f(z)=\frac{z-iy}{z^2-2izy}-\frac{iy}{z^2-2izy}$$
$$f(z)=\frac{z-2iy}{z^2-2izy}=\frac{z-2iy}{z(z-2iy)}=\frac{1}{z}$$
A: Substitute
$$
x=\frac{z+\bar z}{2},\quad y=\frac{z-\bar z}{2\,i}.
$$
All $\bar z$ should desappear, leaving you with an expression deppending only on $z$.
