Differentials (infinitesimals) in complex analysis We have a complex function $ w(z)=w(x+iy)$, and we can write $w(x,y)=u(x,y)+iv(x,y)$.
The derivative is $$\frac{dw}{dz}=\frac{1}{2}(\frac{\partial z}{\partial x}-i\frac{\partial w}{\partial y})$$ (right?)
There are lots of other ways to write this using the Cauchy-Riemann equation.
Now I want to understand the relation between the differentials $dz$, $dx$, $dy$.
I came across this equation: $dz\, d\bar{z}=dx\, dy$. But I cannot figure out why it would be true (or if it is?). I've tried substituting lots of stuff from the Cauchy-Riemann equations, but it doesn't seem to work out, and I don't really understand how the "algebra" with differentials works. Can anyone shed some light on this issue?
 A: Hint:$$x=\frac{z+\bar z}{2}$$ and$$y=\frac{z-\bar z  }{2i}$$ $$\frac{dw}{dz}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial z}$$$$\frac{dw}{dz}=(\frac{1}{2}\frac{\partial z}{\partial x})(\frac{-i}{2}\frac{\partial w}{\partial y})=\frac{1}{2}(\frac{\partial z}{\partial x}-i\frac{\partial w}{\partial y})$$
A: I think what is more of interest here is the relation
$$y \, dx - x \, dy = \frac{1}{2 i} (z \, d\bar{z} - \bar{z} \, dz)$$
Also,
$$\frac{d}{dz} = \frac{1}{2} \left (\frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right )$$
$$\frac{d}{d\bar{z}} = \frac{1}{2} \left (\frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right )$$
A: Since none of the other posted answers have said anything about the 2-form in the question:
Because $z = x+iy$, the 1-form $dz$ can be written as $dz = dx +i\,dy$. Similarly, $d\bar z = d(x-iy) = dx - i\,dy$. Hence
\begin{align}
d\bar z \wedge dz &= (dx - i\,dy) \wedge (dx + i\,dy) \\
&= dx \wedge dx - i\,dy\wedge dx  + i\,dx\wedge dy + dy\wedge dy \\
&= 2i\,dx\wedge dy.
\end{align}
which is useful when doing double integrals in complex coordinates. The area form $dx \wedge dy$ can thus be written as $\frac{1}{2i} d\bar z \wedge dz = \frac{i}{2} dz \wedge d\bar z$. One useful consequence is Cauchy's integral formula for $C^1$, not necessarily holomorphic, functions:
$$f(\zeta) = \frac{1}{2\pi i}\int_{\partial\Omega} \frac{f(z)}{z-\zeta}\,dz + \frac{1}{2\pi i} \iint_\Omega \frac{\frac{\partial f}{\partial \bar z}(z)}{z-\zeta}\,dz\wedge d\bar z.$$
(Note that if $f$ is holomorphic, then $\frac{\partial f}{\partial \bar z} = 0$, and the formula reduces to the familiar version of Cauchy's integral formula.)
