Application of chain rule for a complex variable I'm looking at basic definitions in complex analysis, and I can't figure out where a factor of $1/2$ comes from below.  All sources I've found either invoke it without explanation, or derive it after assuming the Cauchy-Riemann conditions.
Given $\quad f(z) = u(x,y) + i v(x,y),\quad z = x + i y $,  
$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x }\frac{\partial x }{\partial z} + \frac{\partial f }{\partial y}\frac{\partial y}{\partial z}$,
Naively solving $z = x + i y$ for $x$ and $y$ and taking the partial gives:
$\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x } - i\frac{\partial f }{\partial y}$.  
But this is different than the correct result by a factor of $1/2$:  
$\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x } - i\frac{\partial f }{\partial y}\right)$.
QUESTION:  Why exactly does the chain rule fail here, and how does one get the correct result without invoking Cauchy-Riemann equations first?  Other illuminating remarks encouraged.
 A: Here is one approach.  Convert from the two variables $x,y$ to the two variables $z, \overline{z}$.  In one direction
$$
z = x + i y,\qquad \overline{z} = x - i y
$$
Solve to get the other direction
$$
x = \frac{z+\overline{z}}{2},\qquad y=\frac{z-\overline{z}}{2i}
$$
Now we need
$$
\frac{\partial{x}}{\partial{z}} = \frac{1}{2},\qquad
\frac{\partial y}{\partial z} = \frac{1}{2i} .
$$
Put these into $\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x }\frac{\partial x }{\partial z} + \frac{\partial f }{\partial y}\frac{\partial y}{\partial z}$ to get
$$
\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x } - i\frac{\partial f }{\partial y}\right)
$$
A: Since the OP mentioned the Cauchy-Riemann equations, I am assuming that $f$ is analytic.
Now, we have both 
$$f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} \tag 1$$
and
$$f'(z)=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}\tag 2$$
Adding both sides of $(1)$ and $(2)$ and dividing by $2$ reveals that
$$\begin{align}
f'(z)&=\frac12\left(\frac{\partial (u+iv)}{\partial x}-i\frac{\partial (u+iv)}{\partial y}\right)\\\\
&=\frac12\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right)
\end{align}$$
as was to be shown!

NOTE:
We can use a method of differentials that are correct and reflect the OP's main question.  To that end, we can write
$$\begin{align}
df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy
\end{align} \tag 3$$
Then, setting $z=x+iy$, we have $x=z-iy$ so that
$$dx=dz-idy \tag4$$
Substituting $(4)$ into $(3)$ yields
$$df=\frac{\partial f}{\partial x}dz+\left(\frac{\partial f}{\partial y}-i\frac{\partial f}{\partial x}\right)dy \tag 5$$
By setting $y=-iz+ix$, we obtain similarly 
$$df=-i\frac{\partial f}{\partial y}dz+\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)dx\tag 6$$
For $f$ to be analytic, $(5)$ and $(6)$ imply that both (i)
$$\begin{align}
\frac{\partial f}{\partial y}-i\frac{\partial f}{\partial x}&=0 \tag 7\\\\
\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}&=0 \tag 7
\end{align}$$
from which $(7)$ gives the Cauchy-Riemann Equations, and (ii)
$$\begin{align}
df&=\frac{\partial f}{\partial x}dz\\\\
&=-i\frac{\partial f}{\partial y}dz
\end{align} \tag 8$$
from which $(8)$ implies that 
$$\begin{align}
\frac{df}{dz}&=\frac{\partial f}{\partial x}\\\\
&=-i\frac{\partial f}{\partial y}\\\\
&=\frac12\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right)
\end{align}$$
which recovers the expected result!
A: Let me try to clarify a confusion that was brought up in the comments, namely that $\partial x/\partial z\neq(\partial z/\partial x)^{-1}$.
Let us consider a smooth function $f:\mathbb R^n\to\mathbb R^n$ that satisfies $f(0)=0$.
Near zero, $f$ can be approximated by its derivative, so that
$$
f(x)\approx Df(0)x.
$$
(The error term is $O(\|x\|^2)$, but let me drop it altogether.)
Suppose $f^{-1}$ exists and is smooth.
For it we have also $f^{-1}(y)\approx Df^{-1}(0)y$, and so
$$
x=f^{-1}(f(x))\approx Df^{-1}(0) Df(0)x
$$
when $\|x\|$ is small.
This can only be true if $Df^{-1}(0) Df(0)$ is the identity matrix, or, in other words, $Df^{-1}(0)=Df(0)^{-1}$.
Matrix inversion cannot be done element by element, so in general we have $\partial f_i/\partial x_j\neq (\partial x_j/\partial f_i)^{-1}$.
The partial derivative $\partial f_i/\partial x_j$ is an element of the matrix $Df(0)$ and $\partial x_j/\partial f_i$ is an element of the matrix $Df^{-1}(0)$.
The same works for functions $\mathbb C^n\to\mathbb C^n$.
In your case, we are looking at the mapping $(x,y)\mapsto(z,\bar z)=(x+iy,x-iy)$.
This mapping is actually linear, so its derivative is easy to compute; it is
$$
A=
\begin{pmatrix}
1&i\\1&-i
\end{pmatrix}.
$$
The inverse of this matrix is
$$
B=
\frac12
\begin{pmatrix}
1&1\\-i&i
\end{pmatrix}.
$$
Now $\partial z/\partial x=a_{11}=1$ but $\partial x/\partial z=b_{11}=\frac12$.
This is why your calculation was off by a factor of two.
A more direct way to see this would be to express $x$ in terms of $z$ and $\bar z$.
From $x=\frac12(z+\bar z)$ you can see that $\partial x/\partial z=\frac12$.
I'm not sure if this is a particularly fruitful way to approach the Cauchy—Riemann operator(s), but it may give some insight.
One problem is that it is not very clear how to interpret $\partial x/\partial z$ as a partial derivative.
