Consider the differential operators
$\dfrac{\partial}{\partial z}$ and $\dfrac{\partial}{\partial \bar{z} }$ defined by
$\frac{\partial}{\partial z} = \frac {1}{2} (\frac{\partial}{\partial x} - \frac{i \partial}{\partial y}$) ; $\frac{\partial}{\partial \bar{z} } = \frac {1}{2} (\frac{\partial}{\partial x} +\frac{i \partial}{\partial y}$).
Prove that if $f$ is holomorphic then $\frac{\partial}{\partial \bar{z} } = 0$.
I start like that Proof: let $f$ be a holomorphic function such that $f = u + iv$ meaning $f(x,y) = u(x,y) + iv(x,y)$ so that means $f$ has a first partial derivatives that satisfy the cauchy riemann equations.
So i am thinking something like that
$\frac{\partial f}{\partial z} = (\frac{\partial u}{\partial x} + \frac{i \partial v}{\partial x}$) ; $\frac{\partial f}{\partial y } = (\frac{\partial u}{\partial y} +\frac{i \partial v}{\partial y}$)
and the cauchy rieman equation is $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} =- \frac{\partial v}{\partial x}$.
Now i really do not know how to go about this problem, i kept asking what exactly am i looking for here. Help anyone.
The fact is indeed a consequence of the CR equations. Note that
$$\frac{\partial f}{\partial \bar z}=\frac12 \frac{\partial (u+iv)}{\partial x}+\frac12 \frac{\partial (iu-v)}{\partial y}=\frac12 \left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\,\frac12 \left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)=0$$
since the last terms in parantheses are zero by the CR equations and $f$ is analytic by assumption. The result implies that no analytic function can have dependence on $\bar z$ (i.e., cannot be a function of $\bar z$.