What Do Wirtinger Derivatives Do?
24 Oct 2021
I was chatting with a friend about complex analysis earlier today, and I realized that I was never really told why we should care about Wirtinger Deritatives, at least in one complex variable^{1}. I figured I would write up a quick blog post about them, and explain how they can help explain (intuitively) which functions are holomorphic.
First, what are they? Wirtinger derivatives are almost always introduced in the following context:
If $f(x,y) = u(x,y) + i v(x,y) : \mathbb{R}^2 \to \mathbb{C}$, we can view $f$ as a complex function by setting $f(z) = f(\mathfrak{Re}(z), \mathfrak{Im}(z))$. It’s then reasonable to ask how the real differentiability of $f$ relates to the (stronger) complex differentiability of $f$.
We find that $f$ is complex differentiable exactly when it satisfies a pair of partial differential equations called the CauchyRiemann Equations:
\[\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \quad \frac{\partial u}{\partial y} =  \frac{\partial v}{\partial x}\]Many sources then introduce the Wirtinger derivatives,
\[\frac{\partial}{\partial z} = \frac{1}{2} \left ( \frac{\partial}{\partial x}  i \frac{\partial}{\partial y} \right )\]and^{2}
\[\frac{\partial}{\partial z^\ast} = \frac{1}{2} \left ( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right ),\]then reformulate the CauchyRiemann equations in the much simpler form:
The benefit of this is that \(\frac{\partial f}{\partial z^\ast} = 0\) is very memorable. The difficult part is that, formally, this doesn’t make much sense. I’ve known many students who were confused by this notation^{3}, since how a function depends on $z$ is intimately related to how it depends on \(z^\ast\). After all, we can recover \(z\) and \(z^\ast\) from each other, so they aren’t obviously independent in the same way $x$ and $y$ are^{4}!
Here, though, I think it’s helpful to take a simpler, more utilitarian stance. Before asking what these derivatives mean, let’s just see what they do. After working with them for a while, you can decide for yourself whether they’re merely a formal tool (which you’re welcome to do away with) or if they have some deeper meaning^{5}.
First, we need to say how to compute with them. Thankfully, they satisfy all the nice formulas you expect!
As a nice exercise, you should show (some subset of) the following:
$\frac{\partial}{\partial z} z = 1$ and $\frac{\partial}{\partial z^\ast} z = 0$
$\frac{\partial}{\partial z} z^\ast = 0$ and $\frac{\partial}{\partial z^\ast} z^\ast = 1$
Both $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial z^\ast}$ are linear (so, e.g. $\frac{\partial}{\partial z}(f + g) = \frac{\partial}{\partial z}f + \frac{\partial}{\partial z}g$)
Both $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial z^\ast}$ satisfy the power rule, product rule, chain rule, and quotient rule
\(\overline{\frac{\partial}{\partial z} f} = \frac{\partial}{\partial z^\ast} \overline{f}\) and \(\overline{\frac{\partial}{\partial z^\ast} f} = \frac{\partial}{\partial z} \overline{f}\)
Notice that, perhaps counterintuitively, $z^\ast$ gets treated like a constant when we use $\frac{\partial}{\partial z}$ (and vice versa)!
Now we’re finally in a place to see how these can help us!
Consider $f(z) = \mathfrak{Re}(z)$, which picks out the real part of $z$. Naively, it seems reasonable that this should be holomorphic. But notice:
\[\frac{\partial}{\partial z^\ast} \mathfrak{Re}(z) = \frac{\partial}{\partial z^\ast} \frac{z + z^\ast}{2} = \frac{1}{2} \neq 0.\]Since a function is holomorphic if and only if $\frac{\partial}{\partial z^\ast} f = 0$, we see that $\mathfrak{Re}(z)$ is not holomorphic after all!
Similarly, let’s look at $z$. In this case, we expect it to not be differentiable, since this fails even in the real case. And as predicted:
\[\frac{\partial}{\partial z^\ast} z = \frac{\partial}{\partial z^\ast} \sqrt{z z^\ast} = \frac{z}{2 \sqrt{z z^\ast}} \neq 0.\]For polynomials, however, we find that they’re all holomorphic! By linearity it suffices to consider $z^n$, and for those we see
\[\frac{\partial}{\partial z^\ast} z^n = 0.\]Informally, then, we see how this can be useful. To check if a function is holomorphic, we find a way to write it in terms of $z$s and $z^\ast$s, and then compute a single derivative. In many cases of interest, there are simply no $z^\ast$s in sight, which means we get holomorphicity “for free”. For instance, notice
\[\frac{\partial}{\partial z^\ast} e^z = \frac{\partial}{\partial z^\ast} \left ( 1 + z + \frac{z^2}{2} + \ldots \right ) = \frac{\partial}{\partial z^\ast} 1 + \frac{\partial}{\partial z^\ast} z + \frac{\partial}{\partial z^\ast} \frac{z^2}{2} + \ldots = 0\]since there are no $z^\ast$s in this expression at all!
This makes it extremely quick to check if a given function is holomorphic in practice.
A genuinely quick post today! I have some longer form stuff in the works, but I’ve been busy hosting for the past little while. It seems I have too many friends who want to visit me!
I don’t have much else to say about Wirtinger derivatives. I would be interested in hearing about other uses or reasons to care. I think they get used for optimization, where we view $f : \mathbb{C} \to \mathbb{C}$ instead as a function $\mathbb{C} \to \mathbb{R}^2$, but I haven’t looked into it at all… Feel free to leave information about this or other uses in the comments!
See you all soon ^_^

Any more than one complex variable scares me senseless, at least at time of writing. I haven’t actually tried to learn any complex analysis in several varaibles, but I’ve heard horror stories, and I know it’s still an area of active research. ↩

Here I’m using \(z^\ast\) to mean the complex conjugate. I’m TAing a class right now and I misinterpreted $\overline{X}$ to mean the closure of a set $X$ in the plane, when in fact the professor meant it to mean the complement of $X$. Because of this experience I’m a bit twitchy when it comes to overloading the overline right now, and $z^\ast$ is pretty unambiguous. ↩

Indeed, I was personally a student confused by this for a while ↩

That said, they are still linearly independent when viewed over $\mathbb{R}$. After all, if $z = x + iy$, then we can recover $x$ and $y$ as
\[x = \frac{z + z^\ast}{2} \quad \text{ and } \quad y = \frac{z  z^\ast}{2i}\]in fact, these motivate the definition of $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial z^\ast}$. Do you see how? ↩

Personally, I’m still agnostic on this point. I currently view them as as useful mnemonic, and not much more. Though as I understand it, they become more useful and meaningful when working with several complex variables, so I’m keeping an open mind as to their “true meaning” until I learn more. ↩