Partial derivative proof of complex numbers Given $z=x+iy$ and $\overline z=x-iy$, prove the following:
$${\partial\over{\partial x}}={\partial\over{\partial z}}+{\partial\over{\partial{\overline z}}}$$
and
$${\partial\over{\partial y}}=i\left({\partial\over{\partial z}}-{\partial\over{\partial{\overline z}}}\right)$$
I have no clue where to begin equations except that it may be related to Cauchy-Riemann equations. Any help as how to begin the proof and some hints to take note of would be of much help.
 A: $z=x+iy$ and $\overline z=x-iy$
Let $f=f(z,\overline z)$ 
Now ${\partial f\over{\partial x}}={\partial f\over{\partial z}} {\partial z\over{\partial{ x}}}+{\partial f\over{\partial{\overline z}}}{\partial {\overline z}\over{\partial{ x}}}={\partial f\over{\partial z}}(1)+{\partial f\over{\partial{\overline z}}}(1)=({\partial \over{\partial z}}+{\partial \over{\partial{\overline z}}})f$
$\implies{\partial\over{\partial x}}\equiv {\partial\over{\partial z}}+{\partial\over{\partial{\overline z}}}$
Similarly, 
${\partial f\over{\partial y}}={\partial f\over{\partial z}} {\partial z\over{\partial{ y}}}+{\partial f\over{\partial{\overline z}}}{\partial {\overline z}\over{\partial{ y}}}={\partial f\over{\partial z}}(i)+{\partial f\over{\partial{\overline z}}}(-i)=i({\partial \over{\partial z}}-{\partial \over{\partial{\overline z}}})f$
$\implies{\partial\over{\partial y}}\equiv i({\partial\over{\partial z}}-{\partial\over{\partial{\overline z}}})$
A: $dz$ = $dx+ i dy$,
$d \bar z = dx - i dy$
By definition, $\partial_z$ is such a combination of $\partial_x$ and $\partial_y$ that $dz(\partial_z) = 1$
Let's check if $\partial_z = (\partial_x - i\partial_y)/2$ fits.
$dx((\partial_x - i\partial_y)/2)+ idy((\partial_x - i\partial_y)/2) = \frac{1}{2} dx(\partial_x) + \frac{1}{2} idy(-i \partial_y) = 1$.
The same way you check that $\partial_{\bar{z}} = (\partial_x - i \partial_y) /2$
Dealing with the differentials is simple, because you can transform one differential into sum of others fast, but the vector fields are a bit harder to operate so I use duality between differential forms and vector fields. Since this duality is one-to-one, it's sufficient just to check that the equality holds.
A: Let $u$ be an arbitrary continuous and differentiable function of $z$ and $\bar{z}$.
Also as $z = x+ iy$ and $\bar{z} = x-iy$,
$x = \frac{z+\bar{z}}{2}$ and $y = \frac{z-\bar{z}}{2i}$
So, 
$\frac{\partial u}{\partial z} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial z}$
$\frac{\partial u}{\partial z} = \frac{\partial u}{\partial x}\frac{1}{2} + \frac{\partial u}{\partial y}\frac{1}{2i}$
or
$$\frac{\partial }{\partial z} = \frac{\partial }{\partial x}\frac{1}{2} - \frac{\partial }{\partial y}\frac{i}{2} \cdots(i)$$
Similarly,
$\frac{\partial u}{\partial \bar{ z}} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial \bar{z}} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \bar{z}}$
$\frac{\partial u}{\partial \bar{z}} = \frac{\partial u}{\partial x}\frac{1}{2} + \frac{\partial u}{\partial y}\frac{-1}{2i}$
or
$$\frac{\partial }{\partial \bar{z}} = \frac{\partial }{\partial x}\frac{1}{2} + \frac{\partial }{\partial y}\frac{i}{2}\cdots(ii)$$
Adding (i) and (ii),
$$\frac{\partial }{\partial z} + \frac{\partial }{\partial \bar{z}} = \frac{\partial }{\partial x}$$
or
$$\frac{\partial }{\partial x} = \frac{\partial }{\partial z} + \frac{\partial }{\partial \bar{z}}$$
Similarly by subtracting , we may find that
$$\frac{\partial }{\partial y} = i\big(\frac{\partial }{\partial z} - \frac{\partial }{\partial \bar{z}}\big)$$
