Complex derivative - how is this result obtained? I am quite new to the field of complex algebra. During my studies, I have come across the following term:
$$(z - \bar{z}) H(z)$$
where $z$ is a complex variable, $\bar{z}$ its complex conjugate, and $H(z)$ is a complex function depending on $z$. The authors take the complex derivative with respect to $z$ and obtain:
$$\frac{\partial}{\partial z}\left( (z - \bar{z}) H(z) \right) = H(z) + z\frac{\partial H(z)}{\partial z} - \bar{z}\frac{\partial H(z)}{\partial z}$$
Now my question: How was this result obtained?

My own approach would look like this:
$$\frac{\partial}{\partial z}\left( (z - \bar{z}) H(z) \right)$$
Pull $H(z)$ inside the paranthesis
$$\frac{\partial}{\partial z}\left( zH(z) - \bar{z}H(z)  \right)$$
And split up the terms
$$\frac{\partial}{\partial z}\left( zH(z)\right) - \frac{\partial}{\partial z}\left(\bar{z}H(z)  \right)$$
Now apply the chain rule $\frac{\partial}{\partial x}a(x)b(x) = a'(x)b(x) + a(x)b'(x)$:
$$\frac{\partial}{\partial z} zH(z) +  z\frac{\partial}{\partial z}H(z) - \frac{\partial}{\partial z}\bar{z}H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$
Now, I expect $\frac{\partial}{\partial z} z= \frac{\partial}{\partial z} \bar{z} = 1$:
$$H(z) +  z\frac{\partial}{\partial z}H(z) - H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$
which should simplify to:
$$z\frac{\partial}{\partial z}H(z) - \bar{z}\frac{\partial}{\partial z}H(z)$$
That's not the result the authors obtained (the term $H(x)$ is missing). A possible explanation might be that $\frac{\partial}{\partial z} \bar{z}=0$, but I would be surprised if $\bar{z}$ is independent of $z$. Do you know where my mistake is? I have least confidence in the second-to-last step.
 A: Indeed $\frac{\partial}{\partial z} \bar{z} = 0$. Remember that $\frac{\partial}{\partial z}$ is defined as $\frac{\partial}{\partial z} = \frac{\partial}{\partial x} - i\frac{\partial}{\partial y}$,  so $\frac{\partial}{\partial z} \bar{z} = 1 + i^2 = 0$.
A: $\overline z$ is treated as a variable independent of $z$. Of course, it's not really independent, since there's a clearly defined one-to-one correspondence between complex numbers and their conjugates. But some formalisms of complex analysis pretend that it is not so. Which is why I personally dislike those formalisms.
But there is a way to make this rigorous. We can note that if a function $f(x+\mathrm iy)$ is complex differentiable, then $\partial_z f=\partial_x f=-\mathrm i\partial_y f$. From this, we can conclude that
$$\partial_z f=\frac{1}{2}(\partial_x -\mathrm i\partial_y )f.$$
We can also note that we can apply the differential operator on the right to functions which are not complex differentiable. So it's a generalization of the complex derivative operator, and we define $\partial_z:=\frac{1}{2}(\partial_x-\mathrm i\partial_y)$. For complex-differentiable functions, this still only returns the complex derivative. But it will also return something for non-differentiable functions, for instance, $\partial_z\overline z=0$. Using the old meaning where $\partial_z$ is strictly the complex derivative, $\partial_z\overline z$ simply didn't exist, since $\overline z$ is not complex-differentiable. So you wouldn't have been able to use any of the rules for differentiation in the first place.
