# Partial derivative proof of complex numbers

Given $$z=x+iy$$ and $$\overline z=x-iy$$, prove the following: $${\partial\over{\partial x}}={\partial\over{\partial z}}+{\partial\over{\partial{\overline z}}}$$ $$\mathbb {and}$$ $${\partial\over{\partial y}}=i\left({\partial\over{\partial z}}-{\partial\over{\partial{\overline z}}}\right)$$ 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.

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 $$\frac{\partial }{\partial \bar{z}} = \frac{\partial }{\partial x}\frac{1}{2} + \frac{\partial }{\partial y}\frac{i}{2}\cdots(ii)$$

$$\frac{\partial }{\partial z} + \frac{\partial }{\partial \bar{z}} = \frac{\partial }{\partial x}$$ or

$$\frac{\partial }{\partial x} = \frac{\partial }{\partial z} + \frac{\partial }{\partial \bar{z}}$$

Similarly by subtracting , we may find that

$$\frac{\partial }{\partial y} = i\big(\frac{\partial }{\partial z} - \frac{\partial }{\partial \bar{z}}\big)$$

$$z=x+iy$$ 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,

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

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

$$dz$$ = $$dx+ i dy$$, $$d \bar z = dx - i dy$$

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.