Wirtinger Derivatives of Complex Functions
Buildup
Let’s assume the complex function $f : \mathbb{C} \to \mathbb{C}$ is given. The complex number $z=x+iy$, being a linear combination of two real numbers $x,y \in \mathbb{R}$, allows us to consider the function $f$ as a function of two real variables. Moreover, using two real functions $u,v : \mathbb{R}^{2} \to \mathbb{R}$, the value of function $f$ can be divided into real and imaginary parts as follows.
$$ f(z) = f(x,y) = u(x,y) + i v(x,y) $$
Then, the total derivative of $f$ can be expressed as follows.
$$ \begin{equation} \begin{aligned} df &= du + i dv \\ &= \left( \dfrac{\partial u}{\partial x}dx + \dfrac{\partial u}{\partial y}dy \right) + i \left( \dfrac{\partial v}{\partial x}dx + \dfrac{\partial v}{\partial y}dy \right) \\ &= \left( \dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x} \right)dx + \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)dy \end{aligned} \end{equation} $$
From $dz$ and $d\bar{z}$, if we determine $dx, dy$, it can be organized as follows.
$$ \begin{cases} dz = dx + idy \\ d\bar{z} = dx -idy \end{cases} \implies \begin{cases} dx = \dfrac{dz + d\bar{z}}{2} \\ dy = \dfrac{dz - d\bar{z}}{2i} \end{cases} $$
Upon inserting this into $(1)$ and organizing, it can be presented as follows.
$$ \begin{align*} df &= \left( \dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x} \right)dx + \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)dy \\ &= \left( \dfrac{\partial u}{\partial x} + i\dfrac{\partial v}{\partial x} \right)\left( \dfrac{dz + d\bar{z}}{2} \right) + \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)\left( \dfrac{dz - d\bar{z}}{2i} \right) \\ &= \dfrac{1}{2} \left[ \left( \dfrac{\partial u}{\partial x} + i \dfrac{\partial v}{\partial x} \right) -i \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)\right]dz \\ &\quad+ \dfrac{1}{2} \left[ \left( \dfrac{\partial u}{\partial x} + i \dfrac{\partial v}{\partial x} \right) + i \left( \dfrac{\partial u}{\partial y} + i\dfrac{\partial v}{\partial y} \right)\right]d\bar{z} \\ &= \dfrac{1}{2} \left[ \dfrac{\partial (u+iv)}{\partial x} - i \dfrac{\partial (u+iv)}{\partial y}\right]dz + \dfrac{1}{2} \left[ \dfrac{\partial (u+iv)}{\partial x} + i \dfrac{\partial (u+iv)}{\partial y}\right]d\bar{z} \\ &= \dfrac{1}{2} \left[ \dfrac{\partial f}{\partial x} - i \dfrac{\partial f}{\partial y}\right]dz + \dfrac{1}{2} \left[ \dfrac{\partial f}{\partial x} + i \dfrac{\partial f}{\partial y}\right]d\bar{z} \\ &= \dfrac{1}{2} \left[ \dfrac{\partial }{\partial x} - i \dfrac{\partial }{\partial y}\right]fdz + \dfrac{1}{2} \left[ \dfrac{\partial }{\partial x} + i \dfrac{\partial }{\partial y}\right]fd\bar{z} \end{align*} $$
At this point, if we denote the first term’s constant and parenthesis as $\dfrac{\partial }{\partial z}$, and the second term’s parenthesis as $\dfrac{\partial }{\partial \bar{z}}$, we can naturally express the total derivative of the complex function $f$ as follows.
$$ df = \dfrac{\partial f}{\partial z}dz + \dfrac{\partial f}{\partial \bar{z}}d\bar{z} $$
Definition
Let’s define $x,y \in \mathbb{R}$, $z=x+iy$. Assume the complex function $f :\mathbb{C} \to \mathbb{C}$ is expressed by the real function $u,y : \mathbb{R}^{2} \to \mathbb{R}$ as $f(z)=f(x,y)=u(x,y)+iv(x,y)$. The differential operators $\dfrac{\partial }{\partial z}$, $\dfrac{\partial }{\partial \bar{z}}$ are defined as follows.
$$ \begin{align*} \dfrac{\partial }{\partial z} & := \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} - i \dfrac{\partial }{\partial y}\right) \\ \dfrac{\partial }{\partial \bar{z}} & := \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} + i \dfrac{\partial }{\partial y}\right) \end{align*} $$
This is referred to as the (conjugate) Wirtinger differential operator, and $\dfrac{\partial f}{\partial z}, \dfrac{\partial f}{\partial \bar{z}}$ is called the (conjugate) Wirtinger derivative.
Explanation
When applying the Wirtinger differential operator to $z$, $\bar{z}$, the results are as follows.
$$ \dfrac{\partial z}{\partial z} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} - i \dfrac{\partial }{\partial y}\right)(x+iy) = \dfrac{1}{2} \left( \dfrac{\partial x}{\partial x} + \dfrac{\partial y}{\partial y}\right) = 1 $$
$$ \dfrac{\partial \bar{z}}{\partial \bar{z}} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} + i \dfrac{\partial }{\partial y}\right)(x-iy) = \dfrac{1}{2} \left( \dfrac{\partial x}{\partial x} + \dfrac{\partial y}{\partial y}\right) = 1 $$
$$ \dfrac{\partial z}{\partial \bar{z}} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} + i \dfrac{\partial }{\partial y}\right)(x+iy) = \dfrac{1}{2} \left( \dfrac{\partial x}{\partial x} - \dfrac{\partial y}{\partial y}\right) = 0 $$
$$ \dfrac{\partial \bar{z}}{\partial z} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} - i \dfrac{\partial }{\partial y}\right)(x-iy) = \dfrac{1}{2} \left( \dfrac{\partial x}{\partial x} - \dfrac{\partial y}{\partial y}\right) = 0 $$
From these results, the Wirtinger differential operator can be interpreted as treating $z$ and $\bar{z}$ as if they were independent variables. Indeed, since $f$ is non-differentiable for all $z$, $\dfrac{df}{dz}$ does not exist. However, by definition, the Wirtinger operator can differentiate functions including $\bar{z}$, making it natural to refer to the result as a differentiation. It is said that in complex geometry, the Wirtinger differentiation is primarily handled because when examining regular functions its meaning is entirely the same as the traditional meaning of differentiation.
Regarding Regular Functions
Let’s say $f : \mathbb{C} \to \mathbb{C}$ is a regular function. Then, the Wirtinger derivative is as follows.
$$ \begin{align*} \dfrac{\partial f}{\partial z} &= \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} -i \dfrac{\partial }{\partial y}\right) (u + iv) \\ &= \dfrac{1}{2} \left( u_{x} + v_{y} + i(-u_{y} + v_{x}) \right) \end{align*} $$
A differentiable complex function satisfies the Cauchy-Riemann equations, so the equation above is as follows.
$$ \dfrac{\partial f}{\partial z} = \dfrac{1}{2} \left( u_{x} + u_{x} + i(v_{x} + v_{x}) \right) = u_{x} + i v_{x} $$
However, since the derivative of a complex function is $f^{\prime} = \dfrac{df}{dz} = u_{x} + iv_{x}$, the following equation holds.
$$ \dfrac{\partial f}{\partial z} = u_{x} + i v_{x} = \dfrac{df}{dz} $$
Therefore, for the differentiable function $f$, $\dfrac{\partial }{\partial z}$ perfectly matches the meaning of $\dfrac{d}{dz}$. Now, let’s solve equation $\dfrac{\partial f}{\partial \bar{z}}=0$.
$$ \begin{align*} \dfrac{\partial f}{\partial \bar{z}} &= \dfrac{1}{2} \left( \dfrac{\partial }{\partial x} +i \dfrac{\partial }{\partial y}\right) (u + iv) \\ &= \dfrac{1}{2} \left( u_{x} - v_{y} + i(u_{y} + v_{x}) \right) \\ &= 0 \end{align*} $$
As both the real and imaginary parts must be $0$, the following equation is obtained.
$$ \implies \begin{cases} u_{x} = v_{y} \\ u_{y} = - v_{x} \end{cases} $$
This is identical to the Cauchy-Riemann equations. Hence, the equation $\dfrac{\partial f}{\partial \bar{z}}=0$ itself implies that $f$ satisfies the Cauchy-Riemann equations. In other words, the following propositions are all equivalent:
- $f$ is regular (analytic).
- $\dfrac{\partial f}{\partial \overline{z}} = 0$
- $f$ does not depend on $\overline{z}$.
Regarding Non-Regular Functions
Let’s consider functions that include $\overline{z}$, for example, the absolute value $f(z) = \left| z \right|$. Suppose we wish to know the rate of change to optimize such functions. However, since $f$ is non-differentiable, $\dfrac{df}{dz}$ cannot be calculated, and how to optimize $f$ remains unknown. In this case, using the Wirtinger differentiation allows the calculation of something that could be called a gradient.
$$ \dfrac{\partial f}{\partial z} = \dfrac{\partial z\overline{z}}{\partial z} = \overline{z} $$
Indeed, this technique is used in engineering fields such as communications.
Properties
The properties that are rightfully expected to be held by differentiation are well maintained.
Linearity:
$$ \dfrac{\partial (af + g)}{\partial z} = a\dfrac{\partial f}{\partial z} + \dfrac{\partial g}{\partial z} $$
$$ \dfrac{\partial (af + g)}{\partial \overline{z}} = a\dfrac{\partial f}{\partial \overline{z}} + \dfrac{\partial g}{\partial \overline{z}} $$
Product Rule:
$$ \dfrac{\partial (fg)}{\partial z} = \dfrac{\partial f}{\partial z}g + f\dfrac{\partial g}{\partial z} $$
$$ \dfrac{\partial (fg)}{\partial \overline{z}} = \dfrac{\partial f}{\partial \overline{z}}g + f\dfrac{\partial g}{\partial \overline{z}} $$
Chain Rule
$$ \dfrac{\partial (f \circ g)}{\partial z} = \dfrac{\partial f}{\partial w} \dfrac{\partial g}{\partial z} + \dfrac{\partial f}{\partial \overline{w}} \dfrac{\partial \overline{g}}{\partial z} $$
$$ \dfrac{\partial (f \circ g)}{\partial \overline{z}} = \dfrac{\partial f}{\partial w} \dfrac{\partial g}{\partial \overline{z}} + \dfrac{\partial f}{\partial \overline{w}} \dfrac{\partial \overline{g}}{\partial \overline{z}} $$
Additionally,
$$ \dfrac{\partial ^{2} f}{\partial z \partial \overline{z}} = \dfrac{1}{4} \left( \dfrac{\partial ^{2} f}{\partial x^{2}} + \dfrac{\partial ^{2} f}{\partial y^{2}}\right) = \dfrac{1}{4}\Delta f $$
$$ \dfrac{d (f \circ \phi)}{d t} = \dfrac{\partial f}{\partial z} \dfrac{\partial \phi}{\partial t} + \dfrac{\partial f}{\partial \overline{z}} \dfrac{\partial \overline{\phi}}{\partial t} $$
At this point, $\phi : \mathbb{R} \to \mathbb{C}$ is.