Intuition behind derivative with respect to conjugate

Question

When f is a holomorphic function, we have:

$$\frac{df}{d\bar{z}}=0$$

I know how to prove it, but is there some intuition behind this? Like a geometric one or something of those sorts?

Answer 1

Let $f=u+iv$ be holomorphic. We have $$ \frac{\partial f}{\partial \bar{z}} = \frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right) $$ $$ =\frac12\left(u_x+iv_x+i\left(u_y+iv_y\right)\right) $$$$ =\frac12\left(u_x-v_y+i(v_x+u_y)\right). $$ Since $f$ is hoomorphic, $u,v$ satisfy C-R equations. Applying this to the last expression above, we get $\frac{\partial f}{\partial \bar{z}}=0.$ Geometrically change in f happens only if change in $z.$ But w.r.t $\overline{z}$ $f$ continues to remain as constant.

Answer 2

For a more algebraic motivation: Essentially $\frac{df}{d\bar{z}}$ is the complex-antilinear part of $df$.

More precisely, if $f$ is differentiable at $p$ then $df(p):\mathbb R^2\to\mathbb R^2$ is an $\mathbb R$- linear map and then $f$ is complex differentable at $p$ if and only if $$df(p):\mathbb C\to\mathbb C$$

is complex linear (Cauchy-Riemann equations).

Now in general a $\mathbb R$-linear map $T:V\to W$ between complex vectorspaces uniquely decomposes as $T=L+A$, where $L$ is complex linear and $A$ is complex-antilinear (which means $\mathbb R$-linear and $A(iv)=-iA(v)$ for all $v\in V$). Concretely: $$L(v)=\frac12(T(v)-iT(iv))\space \space \space A(v)=\frac12(T(v)+iT(iv))$$

In the complex plane $\mathbb C$ the complex linear maps are of the form $z\mapsto c\cdot z$ and the complex antilinear maps are of the form $z\mapsto c\cdot\bar z$, for some $c\in\mathbb C$, so since

$$df(p)(z)=\frac{df}{d{z}}(p)\cdot z+\frac{df}{d\bar{z}}(p)\cdot \bar z$$

the map $z\mapsto\frac{df}{d\bar{z}}(p)\cdot \bar z$ is the complex-antilinear part of $df(p)$ and so $f$ is holomorphic iff $\frac{df}{d{\bar z}}=0$.