What is the derivative of $z^{-1}$ with respect to $\bar{z}$? I asked a question here a few days ago but it wasn't answered and, as often happens with me, in trying to answer it myself I just confused myself out of understanding what I thought I knew. What is the derivative of $z^{-1}$ with respect to $\bar{z}$ ie $\frac{\partial z^{-1}}{\partial \bar{z}}$. Does such a question even make sense?
More specifically I have a one form on the complex place without the origin $z^{-1}dz$ and wish to get a two form by taking the exterior derivative. This form is clearly not closed so I should get a non zero two form... but what is it?
 A: If $z \neq 0$, put $z = x+iy$. Then $\frac{1}{z} = \frac{x}{x^2+y^2} -i \frac{y}{x^2+y^2}.$ Then, by definition $$\frac{\partial z^{-1}}{\partial \bar{z}} = \frac{1}{2} \left(\frac{\partial z^{-1}}{\partial{x}} +i \frac{\partial z^{-1}}{\partial{y}}  \right).$$ Now, by simple computations one has $$\frac{\partial z^{-1}}{\partial{x}} = \frac{y^2-x^2}{(x^2+y^2)^2} +i \frac{2xy}{(x^2+y^2)^2}$$ and $$\frac{\partial z^{-1}}{\partial{y}} = \frac{-2xy}{(x^2+y^2)^2}+i \frac{y^2-x^2}{(x^2+y^2)^2}.$$ Hence $$\frac{\partial z^{-1}}{\partial \bar{z}} = 0.$$ 
A: You have (at least) two ways.
FIRST: observe that by definition
$$
\frac{\partial}{\partial \bar z}:=
\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)
$$
then express $z^{-1}$ wrt its real and imaginary parts,
$$
z^{-1}=\frac1z=\frac{\bar z}{z\bar z}=\frac{\bar z}{|z|^2}
=\frac{x}{x^2+y^2}-i\frac y{x^2+y^2}
$$
and from what I've just wrote, you have simply to apply the operator
$
\frac{\partial}{\partial \bar z}
$
(called Wirtinger operator) and derive as you are used to do:
\begin{align*}
\frac{\partial}{\partial \bar z}z^{-1}
=&\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)\left[\frac{x}{x^2+y^2}-i\frac y{x^2+y^2}\right]\\
=&\frac12\left[\frac{x^2+y^2-2x^2}{(x^2+y^2)^2}+i\frac{2xy}{(x^2+y^2)^2}-
i\frac{2xy}{(x^2+y^2)^2}+\frac{x^2+y^2-2y^2}{(x^2+y^2)^2}\right]\\
=&0
\end{align*}
SECOND: you have to think at $\bar z$ as a new variable, different from $z$; precisely $\bar z$ is the variable conjugate to $z$.
As such, you have to express $z^{-1}$ wrt $z$ and $\bar z$, i.e. $z^{-1}=z^{-1}(z,\bar z)$: 
$$
z^{-1}=\frac1z
$$
in particular it is constant wrt $\bar z$, thus deriving classically wrt the $\bar z$ you get $0$:
\begin{align*}
\frac{\partial}{\partial \bar z}z^{-1}
=&\frac{\partial}{\partial \bar z}\frac1{z}=0\;\;.
\end{align*}
