How to compute $\frac{\partial f}{\partial \overline{z}}$ to show holomorphicity I have recently learned in my complex variables class to check if a function is holomorphic by verifying the following equation, which is equivalent to the Cauchy-Riemann equations. $$\frac{\partial f}{\partial \overline{z}}=0$$ I have found the following example in my textbook. We check that $f(z)=z^2-z$ is holomorphic as follows: $$\frac{\partial f}{\partial \overline{z}}=\frac{\partial}{\partial\overline{z}}(z^2-z)=2z\frac{\partial z}{\partial \overline{z}}-\frac{\partial z}{\partial \overline{z}}=0-0=0$$This might seem like an extremely basic question, but I do not understand the step after the second equals sign. Why are they taking derivatives with respect to $z$ first? Is it not obvious that the derivative is $0$ simply from the fact that there are no $\overline{z}$ in $f(z)$? Neither my notes nor the textbook explain why that extra step occurs. 
I would appreciate any clarification.
 A: By definition, we have that 
\begin{align}
\frac{\partial}{\partial \bar z} = \frac{1}{2} \left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y} \right)
\end{align}
and by simply calculation we also have that
\begin{align}
\frac{\partial}{\partial \bar z} \bar z= \frac{1}{2} \left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y} \right)(x-iy) = 1
\end{align}
and
\begin{align}
\frac{\partial}{\partial \bar z} z = \frac{1}{2} \left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y} \right)(x+iy) =0.
\end{align}
Then we have
\begin{align}
\frac{\partial}{\partial \bar z} (z^2-z) = \frac{\partial}{\partial \bar z}z^2 - \frac{\partial}{\partial \bar z}z = \frac{\partial}{\partial \bar z}(z) z+ z\frac{\partial}{\partial \bar z}(z) -\frac{\partial}{\partial \bar z}z = 0\cdot z+z\cdot 0 -0 = 0  
\end{align}
where the product rule and linearity come from the fact that $\partial_x$ and $\partial_y$ are linear and satisfy the product rule. 
A: Let's compute this:
$$
\frac{\partial}{\partial q}(z^2)
$$
How would you do that?  I would do
$$
2 z \frac{\partial z}{\partial q}
$$
Would you call that "derivative with respect to $z$ first"?  
How about:
$$
\frac{d}{dt}\;\cos^2 t = 2\cos t \;\frac{d}{dt}\;\cos t
$$
Would you call that "derivative with respect to $\cos t$ first"?  
These examples are the same principle as in your question.
