On the derivative with respect to $\bar z$

Question

There are similar questions to this one, but unfortunately they do not fully address my issue. I was given the following definition of a Wirtinger operator: $$\frac{\partial}{\partial \bar{z}}:=\frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)$$ First, consider the following example, which will make it easier to understand my trouble.

Let $f:\mathbb{C} \to \mathbb{C}, f(z)=a^3\bar z e^z + z^{10}-\frac{b}{2}\bar z - 5e^z-5i \bar z e^z + 8 \bar z$. Determine the values of $a, b \in \mathbb{C}$ such that $f$ is entire (=holomorphic on the whole complex plane).

The usual way to solve this kind of problem is to observe that $f(z) = u(x,y) + iv(x,y)$ where $u, v \in C^\infty(\mathbb{R^2})$ . Therefore $f$ is holomorphic on $\mathbb{C}$ if and only if the Cauchy-Riemann equations hold for every $ z \in\mathbb{C}$ and this happens if and only if $$\frac{\partial f}{\partial \bar{z}} (z)= 0.$$ Basically, to proceed, one just computes the partial derivative of the expression of $f$ as $\bar z$ were a variable (independent from $z$), applying the standard rules for derivatives, like the product and the chain rule.

I wonder why it works. Let me be clearer. Suppose that $\bar z$ appears in the expression of $f$. Then, it is natural to associate to $f$ the function $g:\mathbb{C^2} \to \mathbb{C}$ such that $f(z) = g(z, \bar z)$. Then, consider the following operator acting on differentiable functions: $$O[f](z):=\frac{1}{2}\left(\frac{\partial f}{\partial x} (x,y) + i\frac{\partial f}{\partial y}(x,y)\right)$$ where $z=x+iy=(x,y)$. As discussed earlier, the problem is "simplified" to $O[f](z)=0$.

But why does $O[f](z) = \frac{\partial g}{\partial \bar{z}} (z)$?

My question is motivated by the fact that it is really easy to compute $\frac{\partial g}{\partial \bar{z}}$ in the example I suggested above: in fact, we have $$0=\frac{\partial g}{\partial \bar{z}} (z) = a^3 e^z -\frac{b}{2}-5ie^z+8$$

but it looks really complicated to use the actual definition of $O$. I see that the expression for $\frac{\partial f }{\partial \bar{z}}$ can be simplified using the linearity of $O$, which easily follows from that of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$, so that $$0=\frac{\partial f }{\partial \bar{z}} (z)= \frac{\partial}{\partial \bar{z}} \left(a^3\bar z e^z + z^{10}-\frac{b}{2}\bar z - 5e^z-5i \bar z e^z + 8 \bar z \right) = \frac{\partial}{\partial \bar{z}} \left(a^3\bar z e^z -\frac{b}{2}\bar z -5i \bar z e^z + 8 \bar z \right)$$

since $z^{10}$ and $-5e^z$ are holomorphic and thus their derivative with respect to $\bar z$ is null. However, I can't see how one would proceed without the trick (which currently I find unjustified) to proceed as I illustrated above. It looks much more complicated if there is a function of $\bar z$ in the expression of $f$: consider, for example $$f(z)=z^2+z\cos(3\bar z)+\exp(\bar z).$$ The trick gives immediately $$\frac{\partial g}{\partial \bar{z}} (z)= -3\sin(3 \bar z) + \exp(\bar z)$$ while applying the definition of $O$ requires a huge amount of time.

Answer 1

If $f(z) = g(z, \bar z)$ then $$\frac{\partial f}{\partial x}=\frac{\partial g}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial g}{\partial\bar z}\frac{\partial\bar z}{\partial x}=\frac{\partial g}{\partial z}+\frac{\partial g}{\partial\bar z}$$ and similarly, $$\frac{\partial f}{\partial y}=i\frac{\partial g}{\partial z}-i\frac{\partial g}{\partial\bar z}$$ hence $$\frac{\partial g}{\partial\bar z}=\frac12\left(\frac{\partial f}{\partial x}+ i\frac{\partial f}{\partial y}\right).$$

Answer 2

(There is a sense in which the following is the same calculation as Anne Bauval did, but I think it may help some people get a more concrete feel for what's going on and/or how they might have confirmed this themselves.)

Suppose $g\left(u,v\right)$ is real-differentiable with partial derivative functions $g_{1}$ and $g_{2}$. Then what it means for "the trick to work" is that: $$\dfrac{\partial}{\partial z}g\left(z,\overline{z}\right)=\dfrac{\partial}{\partial z}g\left(x+iy,x-iy\right)=\left.\dfrac{\partial}{\partial u}g\left(u,v\right)\right|_{u=x+iy,\,v=x-iy}=g_{1}\left(z,\overline{z}\right)$$ $$\text{and }\dfrac{\partial}{\partial\overline{z}}g\left(z,\overline{z}\right)=\dfrac{\partial}{\partial\overline{z}}g\left(x+iy,x-iy\right)=\left.\dfrac{\partial}{\partial v}g\left(u,v\right)\right|_{u=x+iy,\,v=x-iy}=g_{2}\left(z,\overline{z}\right)\text{.}$$

We have:

\begin{align*} \hphantom{=} & \dfrac{\partial}{\partial z}g\left(z,\overline{z}\right)\\ = & \dfrac{1}{2}\left(\dfrac{\partial}{\partial x}-i\dfrac{\partial}{\partial y}\right)g\left(x+iy,x-iy\right)\\ = & \dfrac{1}{2}\left(\dfrac{\partial}{\partial x}g\left(z,\overline{z}\right)-i\dfrac{\partial}{\partial y}g\left(z,\overline{z}\right)\right)\\ = & \dfrac{1}{2}\left(g_{1}\left(z,\overline{z}\right)\dfrac{\partial\left(x+iy\right)}{\partial x}+g_{2}\left(z,\overline{z}\right)\dfrac{\partial\left(x-iy\right)}{\partial x}-i\dfrac{\partial}{\partial y}g\left(z,\overline{z}\right)\right)\\ = & \dfrac{1}{2}\left(g_{1}\left(z,\overline{z}\right)+g_{2}\left(z,\overline{z}\right)-i\dfrac{\partial}{\partial y}g\left(z,\overline{z}\right)\right)\\ = & \dfrac{1}{2}\left(g_{1}\left(z,\overline{z}\right)+g_{2}\left(z,\overline{z}\right)-i\left(g_{1}\left(z,\overline{z}\right)\dfrac{\partial\left(x+iy\right)}{\partial y}+g_{2}\left(z,\overline{z}\right)\dfrac{\partial\left(x-iy\right)}{\partial y}\right)\right)\\ = & \dfrac{1}{2}\left(g_{1}\left(z,\overline{z}\right)+g_{2}\left(z,\overline{z}\right)-i\left(ig_{1}\left(z,\overline{z}\right)-ig_{2}\left(z,\overline{z}\right)\right)\right)\\ = & \dfrac{1}{2}\left(g_{1}\left(z,\overline{z}\right)+g_{2}\left(z,\overline{z}\right)+g_{1}\left(z,\overline{z}\right)-g_{2}\left(z,\overline{z}\right)\right)\\ = & \dfrac{1}{2}\left(2g_{1}\left(z,\overline{z}\right)\right)\\ = & g_{1}\left(z,\overline{z}\right)\checkmark \end{align*}

Similarly: \begin{align*} \hphantom{=} & \dfrac{\partial}{\partial\overline{z}}g\left(z,\overline{z}\right)\\ = & \dfrac{1}{2}\left(\dfrac{\partial}{\partial x}\boldsymbol{+}i\dfrac{\partial}{\partial y}\right)g\left(x+iy,x-iy\right)\\ = & \cdots\\ = & \dfrac{1}{2}\left(g_{1}\left(z,\overline{z}\right)+g_{2}\left(z,\overline{z}\right)\boldsymbol{+}i\left(ig_{1}\left(z,\overline{z}\right)-ig_{2}\left(z,\overline{z}\right)\right)\right)\\ = & \dfrac{1}{2}\left(g_{1}\left(z,\overline{z}\right)+g_{2}\left(z,\overline{z}\right)-g_{1}\left(z,\overline{z}\right)+g_{2}\left(z,\overline{z}\right)\right)\\ = & \dfrac{1}{2}\left(2g_{2}\left(z,\overline{z}\right)\right)\\ = & g_{2}\left(z,\overline{z}\right)\checkmark \end{align*}