geometric meaning of differentiation with respect to the complex conjugate of $z$

Question

The derivative of a function $f: C \rightarrow C$ with respect to $\overline{z}$ is defined as $$ \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) $$ where $\frac{\partial}{\partial x}$ is defined as the operator which maps $u+iv$ to $u_x + iv_x$, and $\frac{\partial}{\partial y}$ is defined likewise. I am wondering whether there is a geometric way to think about this quantity.

Answer 1

Geometrically one can see what's going on by analyzing $f$'s induced map on tangent vectors. You'll see that the operator $\frac{\partial f}{\partial \overline{z}}$ is measuring the failure of a function to respect multiplication by $i$ between domain tangent vectors and range tangent vectors. It is helpful to think of $f$ mapping from a domain $C$ ~ $R^2$ to a distinct range $C$ ~ $R^2$. Any differentiable function $f$ will induce a map from tangent vectors in the domain to tangent vectors in the range, denoted here $v \rightarrow f'(v)$. This is the derivative of $f$ along the vector $v$.

Both our domain and range are complex and thus have a concept of multiplication by $i$ on tangent vectors. Multiplying by $i$ is like rotating a vector counter-clockwise 90 degrees, e.g. at a point $z=(x,y)$, the vector $\frac{\partial}{\partial x}$ gets sent to $\frac{\partial}{\partial y}$ under multiplication by $i$. The beauty of holomorphic functions is that they respect this 90 degree rotation between the domain and range. That is, if $f$ is holomorphic, then $f'(iv) = i f'(v)$ where the first $i$ multiplication is in the domain and the second represents $i$ multiplication in the range. So, from the image of a vector in the domain (the derivative in one direction at some point), you can infer the image of $f'$ in every direction at that point.

We write this as, for all $v$, $f'(iv) = if'(v)$. The failure to achieve holomorphicity is thus $f'(iv)-if'(v)$ being different from $0$ for some $v$. The operator $\frac{\partial f}{\partial \overline{z}}$ is essentially this measurement for a specific v. Let v be the unit vector $\frac{-\partial}{\partial y}$. Then $i v$ is $\frac{\partial}{\partial x}$. The above formula $f'(iv)-if'(v)$ becomes $f'(\frac{\partial}{\partial x}) - i f'(-\frac{\partial}{\partial y})$ = $\frac{\partial f}{\partial x} + i (\frac{\partial f}{\partial y})$ = $2 \frac{\partial f}{\partial \overline{z}}$.

Of course there's a certain symmetry here. While functions where $\frac{\partial f}{\partial \overline{z}}=0$ are holomorphic, functions where $\frac{\partial f}{\partial z}=0$ are anti-holomorphic; $f'$ sends the vector $i v$ to $- i f'(v)$ for all $v$.

Answer 2

For getting geometric intuition about complex analysis Tristram Needham's book "Visual Complex Analysis" seems to be the best place you can turn to!

There, reading from page 216 on (and you can safely just jump there, if you already have arrived at questions like the one you are asking here), you will learn that holomorphic functions are the ones which look locally like compositions of rotations, translations and rescalings (I mean if you really draw the complex plane as a plane), i.e. they are the ones which carry disks to disks (and I mean the real round disks, with boundary a circle :-).

So since the derivation of a function $f$ with respect to $\bar{z}$ is zero iff $f$ is holomorphic, you could say that it measures how far $f$ is from being holomorphic or, in view of the above characterization, how much a disk can get deformed (e.g. into an ellipsoid) when you apply $f$.

Answer 3

The dee zee bar operator in the secret blogging seminar: What´s up with dee zee bar

Answer 4

This is a rephrasing of aspects of what several people above have said. In particular Weaam's 3rd paragraph shows you how to prove:

A $\mathbb R$-linear function $f : \mathbb R^2 \to \mathbb R^2$ can be decomposed uniquely into a sum $f = Cf+Af$ of two $\mathbb R$-linear functions $Cf, Af : \mathbb R^2 \to \mathbb R^2$, where $Cf$ is "complex linear" meaning that $Cf(av+bw)=aCf(v)+bCf(w)$ for all complex numbers $a,b$ and vectors $v,w\in \mathbb R^2$. $Af$ is complex anti-linear $Af(av+bw)=\overline{a}Af(v)+\overline{b}Af(w)$.

So given an arbitrary Frechet ($\mathbb R$-differentiable) function $f$ defined on an open subset of $\mathbb R^2$, $\partial f/\partial z$ is the complex-linear part of the derivative of $f$. $\partial f/\partial \overline{z}$ is the complex anti-linear part of the derivative of $f$. So this explains why $f$ is complex differentiable if and only if $\partial f/\partial \overline{z} = 0$. In the notation of the previous paragraph, $\partial f/\partial \overline{z} = Af'$, and $\partial f/\partial z = Cf'$.

Answer 5

I think the hypothesis that f is a totally differentiable function is necessary.

A basic first result is that, f is holomorphic if and only if $\frac{\partial f}{\partial \overline{z}}=0$, and in that case $f' = \frac{\partial f}{\partial z}$, i.e "Holomorphic functions are independent of $\overline{z}$ and depend only on z", by Cauchy-Riemann.

Interpreting f as a real function of two independent real variables x, y, where z = x + i y. Hence $x = \frac{z + \overline{z}}{2}$, and $y = \frac{z - \overline{z}}{2i}$

Minimizing f with respect to $\overline{z}$ and applying the chain rule

$\frac{\partial f}{\partial \overline{z}} = \frac{\partial x}{\partial \overline{z}}_z\frac{\partial f}{\partial x}_y + \frac{\partial y}{\partial \overline{z}}_z\frac{\partial f}{\partial y}_x$ $= \frac{1}{2} ( \frac{\partial f}{\partial x}_x + i \frac{\partial f}{\partial y}_y)$ and to minimize over all z = x + iy (since minimizing w.r.t x (resp. y) is to have the partial derivative = 0), then $\frac{\partial f}{\partial \overline{z}} = 0$. see http://people.ccmr.cornell.edu/~muchomas/P480/Notes/dft/node10.html

It should be mentioned that both $\frac{\partial f}{\partial \overline{z}}$ along with $\frac{\partial f}{\partial z}$ are known as Wirtinger derivatives(operators). I'm not sure if geometric intuition is necessary, but these derivatives give the Wirtinger calculus. One builds a theory (a "symbolical tool"?!!) that's analogous to the real case that is applied even to nonholomorphic functions (see "Theory of complex functions", Remmert, page 67-69 for a quick development), or the beginnings of "The Complex Gradient Operator and the CR-Calculus, ECE275A – Lecture Supplement", which is more elementary but could be useful.

Answer 6

This goes along with the question of how to interpret $$ \frac{\partial}{\partial z} \;=\; \frac{1}{2}\left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right) $$ for non-holomorphic functions.

One possible answer is that $\partial f / \partial z$ is the average value of $$ \frac{f(w) - f(z)}{w - z} $$ as $w$ takes values in a small circle centered at $z$. To be precise: $$ \frac{\partial f}{\partial z}(z) \;=\; \lim_{r\to 0}\frac{1}{2\pi}\int_0^{2\pi} \frac{f(z+re^{it}) - f(z)}{re^{it}} dt. $$ Similarly, $\partial f / \partial \overline{z}$ is the average value of $$ \frac{f(w) - f(z)}{\overline{w} - \overline{z}} $$ as $w$ takes values in a small circle centered at $z$.

Answer 7

The formulas $$\frac{\partial f}{\partial z} \;=\; \frac{1}{2}\left(\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y}\right)$$ $$\frac{\partial f}{\partial \bar{z}} \;=\; \frac{1}{2}\left(\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y}\right)$$ are just the trigonometric-to-exponential Fourier coefficient transformations $$c_1=\frac{1}{2}(a_1-ib_1)$$ $$c_{-1}=\frac{1}{2}(a_1+ib_1)$$