TITLE: What is the intuition behind the Wirtinger derivatives?
QUESTION [46 upvotes]: The Wirtinger differential operators are introduced in complex analysis to simplify differentiation in complex variables. Most textbooks introduce them as if it were a natural thing to do. However, I fail to see the intuition behind this. Most of the time, I even think they tend to make calculations harder.
Is there a simple interpretation of these operators? What mental picture do you have when you use them?

REPLY [5 votes]: In the first place, as noted above: the Wirtinger partial derivatives "act as they should''. The proof is not trivial, but it is easily seen with simple examples. Not only the two examples above, below the cited phrase, but also somewhat more complicated: 
$$    \frac{\partial z\overline{z}}{\partial z} = \overline{z} $$    and 
$$    \frac{\partial z\overline{z}}{\partial \overline{z}} = z.$$ For instance:
$$       \frac{\partial z\overline{z}}{\partial z} = \frac{1}{2}\left(
                     \frac{\partial \, x^2+y^2}{\partial x}
                         -   i  \frac{\partial \, x^2+y^2}{\partial y}\right)
         = \frac{1}{2}\left(2x-   i  2y\right) = x- i  y = \overline{z}$$
Of course, the whole point of the Wirtinger partial derivatives is their behavior as ordinary partial derivatives, with a jump immediately from $\partial z\overline{z}/\partial z$ to  $\overline{z}$. This is my "mental picture".
In the second place, this mental picture of ordinary partial derivatives can be applied to prove the rule that a function $f$ is complex-differentiable with respect to $z$ if and only if ${\partial f}/{\partial \overline{z}} = 0$, as follows. Start with
\begin{align*}
\frac{\mathrm{d} f}{\mathrm{d} z} &= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \,
           \frac{\mathrm{d} \overline{z}}{\mathrm{d} z}\\
&= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \,
 \lim_{\Delta z  \rightarrow 0} \frac{\Delta \overline{z} }{\,\Delta z \,}\\
&= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \,
 \lim_{\Delta z  \rightarrow 0} \frac{\overline{\Delta z} }{\,\Delta z \,}\\  
&= \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \,
 \lim_{\Delta z  \rightarrow 0} e^{- i   2\arg(\Delta z  )}\\
& = \frac{\partial f}{\partial z} +
 \lim_{\Delta z  \rightarrow 0}  \left( \frac{\partial f}{\partial \overline{z}} \,
      e^{- i   2\arg(\Delta z  )} \right).
\end{align*}
Then, if  ${\partial f}/{\partial \overline{z}} = 0$ holds then we have indeed:
$$
       \frac{\mathrm{d} f}{\mathrm{d} z} = \frac{\partial f}{\partial z} +
     \lim_{\Delta z  \rightarrow 0} \left( 0 \times  e^{- i   2\arg(\Delta z  )} \right)
    =  \frac{\partial f}{\partial z} +
    \lim_{\Delta z  \rightarrow 0} 0 = \frac{\partial f}{\partial z}
$$
Otherwise, indeed, ${\mathrm{d} f}/{\mathrm{d} z}$ does not exist, because then the $\lim_{\Delta z  \rightarrow 0}$ does not exist.