Partial derivative proof of complex numbers

Question

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.

Answer 1

$z=x+iy$ 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,

${\partial f\over{\partial y}}={\partial f\over{\partial z}} {\partial z\over{\partial{ y}}}+{\partial f\over{\partial{\overline z}}}{\partial {\overline z}\over{\partial{ y}}}={\partial f\over{\partial z}}(i)+{\partial f\over{\partial{\overline z}}}(-i)=i({\partial \over{\partial z}}-{\partial \over{\partial{\overline z}}})f$

$\implies{\partial\over{\partial y}}\equiv i({\partial\over{\partial z}}-{\partial\over{\partial{\overline z}}})$