$\frac{\partial f}{\partial \bar z}\equiv 0$ as a definition for a Holomorphic function?

Question

In (Krantz, 2008; pg14) and in other places I have seen, the definition of a Holomorphic function is one for which: $$\frac{\partial f}{\partial \bar z}\equiv 0$$ rather then the standard definition of $f$ been complex differentiable. My question is are these two definitions equivalent, and how does the former take account of singularities where the function is not analytic?

Answer 1

$f$ is complex differentiable at $a$ iff $$f(a+z) = f(a)+ cz + o(|z|)$$ for some $c \in \mathbb{C}$, thus we can say $ f'(a)= c$.

If it is only real-differentiable then $cz$ is replaced by $M (x,y)$ for some matrix $M \in \mathbb{R}^{2 \times 2}$, equivalently by $c z+\overline{d} \overline{z}$, thus we can say $$\frac{\partial f}{\partial z}(a) = c ,\quad \frac{\partial f}{\partial \bar z}(a) = \overline{d}, \qquad f(a+z) = f(a)+ cz +\overline{d} \overline{z}+ o(|z|)$$

Answer 2

There are at least three equivalent ways to define complex differentiability. Let $f\colon \mathbb C\to\mathbb C$ be a real differentiable function (i.e., differentiable as a function from $\mathbb R^2$ to $\mathbb R^2$). Then $f$ is complex differentiable if any of the three equivalent statements holds:

$(i)$ Cauchy-Riemann equations hold, i.e. if $f(x+iy) = u(x,y)+iv(x,y)$, then

\begin{align} \dfrac{\partial u}{\partial x} &= \dfrac{\partial v}{\partial y},\\ \dfrac{\partial u}{\partial y} &= -\dfrac{\partial v}{\partial x}. \end{align}

$(ii)$ $f(z)$ is independent of $\bar z$, i.e. $\dfrac{\partial f}{\partial\bar z} = 0$.

$(iii)$ Differential of $f$ acts as complex multiplication, i.e. $Df(z) = \lambda z$, $\lambda\in\mathbb C$.

I'm guessing that Cauchy-Riemann equations are "the familiar definition", so let us reduce $(ii)$ and $(iii)$ to $(i)$.

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First of all, when we write $f(x+iy) = u(x,y)+iv(x,y)$, we are thinking of $x$ and $y$ as coordinate functions for which we have corresponding directional derivatives $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}.$ But, we could also look at another coordinates, namely $z = x + iy$ and $\bar z = x - iy$. We would like to represent $\frac{\partial}{\partial \bar z}$ in terms of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$, so first we calculate:

\begin{alignat}{2} \frac{\partial z}{\partial x} &=&\ 1,\ \frac{\partial z}{\partial y} &=&\ i,\\ \frac{\partial \bar z}{\partial x} &=&\ 1,\ \frac{\partial \bar z}{\partial y} &=&-i, \end{alignat}

which gives us

\begin{align} \frac{\partial}{\partial x} &= \frac{\partial}{\partial z} + \frac{\partial}{\partial \bar z},\\ \frac{\partial}{\partial y} &= i\frac{\partial}{\partial z} - i\frac{\partial}{\partial \bar z}, \end{align}

and inverting the system matrix gives us

\begin{align} \frac{\partial}{\partial z} &= \frac 12\left(\frac{\partial}{\partial x} -i \frac{\partial}{\partial y}\right),\\ \frac{\partial}{\partial \bar z} &= \frac 12\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right), \end{align}

and finally

$$2\frac{\partial f}{\partial \bar z} = \left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)(u+iv) = \left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\right) + i\left(\frac{\partial u}{\partial y} + \frac{\partial v }{\partial x}\right)$$

and you can easily see that $(i)$ and $(ii)$ are equivalent.

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Personally, $(iii)$ is my favorite way of looking at complex differentiation and to understand completely its relation to $(i)$, one should look at $\mathbb C$ as $2$-dimensional real algebra.

For now, we have differential of $f$ as real vector function, its matrix representation is the Jacobian

$$J_f=\begin{pmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y}\\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{pmatrix}.$$

Let us examine what does it mean for linear operator on $\mathbb R^2$ to act as complex multiplication. Take some linear operator $A$. We would like that $Az = (a+ib) z$, so and let us see how it acts on basis $\{1,i\}$. $A(1)$ should of course be equal to $a + ib$, while $A(i) = (a+ib)i = -b + ia$. We see that $A$ is represented by matrix $$\begin{pmatrix} a & -b\\ b & a \end{pmatrix}.$$ It is immediate that not only is this necessary, but sufficient as well, since

$$\begin{pmatrix} a & -b\\ b & a \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}= \begin{pmatrix} ax-by\\ bx+ay \end{pmatrix},$$ precisely what we expect of complex multiplication. If you inspect Jacobian again, you will see that $(ii)$ is equivalent to Cauchy-Riemann equations.

What is happening here is that we have faithful representation of real algebra $\mathbb C$

$$\pi\colon \mathbb C\to M_2(\mathbb R),\\ \pi(a+ib) = \begin{pmatrix} a & -b\\ b & a \end{pmatrix}.$$ Moreover, this representation gives $\mathbb R^2$ structure of $\mathbb C$-module isomorphic to $\mathbb C$ with isomorphism $\varphi(a+ib) = (a,b)$. Indeed, $\varphi$ is obviously bijective and $\varphi(\lambda z) = \pi(\lambda)\varphi(z)$.

To sum it up, with real differential we want differential to be $\mathbb R$-linear, while for complex differential we want it to be $\mathbb C$- linear. Since we understand how real differentiation works, we exploit $\mathbb C$-linear isomorphism of $\mathbb C$ and $\mathbb R^2$ to write $\mathbb C$-linearity of complex differential in those terms.