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I know that we can just define the differential operators $$\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})$$ $$\frac{\partial}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y})$$ and that $f$ is holomorphic means that $\frac{\partial f}{\partial \bar{z}}=0$, that's all clear. However, what I don't get it is how we can write any arbitrary function $f(z)=f(z,\bar{z})$, and then calculate 'formally' with the above differential operators and expect the outcomes to be as expected (or even well defined).

So this question is really not about the derivation/definition of the Wirtinger Derivatives, that's all document very well. It's about why we can write for example

\begin{align} \frac{\partial (g\circ f)}{\partial z}&=\frac{\partial (g(f(z,\bar z),\bar f(z,\bar z))}{\partial z}\\\\ &=\left.\frac{\partial g(w,\bar w)}{\partial w}\right|_{w=f(z,\bar z)}\times \frac{\partial f(z,\bar z)}{\partial z}+\left.\frac{\partial g(w,\bar w)}{\partial \bar w}\right|_{\bar w=\bar f(z,\bar z)}\times \frac{\partial \bar f(z,\bar z)}{\partial z}\\\\ &=\left(\frac{\partial g}{\partial z}\circ f\right)\frac{\partial f}{\partial z}+\left(\frac{\partial g}{\partial \bar z}\circ f\right)\frac{\partial \bar f}{\partial z} \end{align}

(copied from here). I just don't see how the formal definition of the Wirtinger derivatives makes it so that all of this goes through.

An answer to this question would explain, rigorously, why the steps in the above calculation are justified, starting from the fact we can write $f(z)=f(z,\bar{z})$ in a well-defined way s.t. the operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ behave as expected.

Thank you

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    $\begingroup$ I don't understand your notation. If $f(z)=f(z,\overline z)$ then $f$ is both a function of one variable and a function of two variables? How does that work? $\endgroup$ – Gregory Grant Feb 19 '16 at 22:38
  • $\begingroup$ > However, what I don't get it is how we can write any arbitrary > function $f(z)=f(z,\bar z)$ $$ $$ You can't. $f(\bar z)$ means you replace your variable $z$ with $\bar z$. $\endgroup$ – Kaster Feb 19 '16 at 22:41
  • $\begingroup$ Yes, that's exactly my confusion, I don't get this, yet people seem to do it. $\endgroup$ – user2520938 Feb 19 '16 at 22:41
  • $\begingroup$ Dr. MV has a great write up already, but I want to add something. ANY function of a variable (say $x$) is also a function of $x$ and $y$, $y$ might just not show up. There are some technical considerations here, but morally this is true. $\endgroup$ – user223391 Feb 20 '16 at 2:22
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Note that the referenced post does not suggest that $f$ is a function of $z$ only and nowhere do we see written "$f(z)=f(z,\bar z)$." Rather, a complex-valued function, $f$, is in general, a function of both $z$ and $\bar z$. To see this, let's take a closer look at things.

Let $\hat f$ be a complex function. Then we can write $\hat f$ in terms of its real and imaginary parts

$$\hat f(x,y)=u(x,y)+iv(x,y) \tag 1$$

where $ u(x,y)$ and $ v(x,y)$ are real=valued functions of $x$ and $y$ with

$$u(x,y)=\text{Re}(\hat f(x,y))$$

and

$$v(x,y)=\text{Im}(\hat f(x,y))$$

Next, note that we can write $x$ and $y$ in terms of $z=x+iy$ and $\bar z=x-iy$ as

$$x=\frac12(z+\bar z) \tag 2$$

and

$$y=\frac{1}{2i}(z-\bar z) \tag 3$$

Substituting $(2)$ and $(3)$ into $(1)$ reveals

$$\begin{align} \hat f(x,y)&= u\left(\frac12(z+\bar z),\frac{1}{2i}(z-\bar z)\right)+i v\left(\frac12(z+\bar z),\frac{1}{2i}(z-\bar z)\right)\\\\ &=f(z,\bar z)\\\\ \end{align}$$

for some function $f$ of $z$ and $\bar z$. So, any complex-valued function that can be expressed as in $(1)$ can be expressed as a function of $z$ and $\bar z$.

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  • $\begingroup$ Very good write up, +1 $\endgroup$ – user223391 Feb 20 '16 at 2:19
  • $\begingroup$ @avid19 Thanks Zach! Much appreciative. - Mark $\endgroup$ – Mark Viola Feb 20 '16 at 4:27
  • $\begingroup$ This still doesn't make sense. we have $u:\mathbb{R}^2\to \mathbb{R}$. However, if we plug in $f(i,0)=u(\frac{1}{2}i,\frac{1}{2})+v(...)$ we see this is not defined. Hence this does not answer the question, because this is still just playing with notation. $f(z,\bar{z})$ does not behave as a 'usual' function in 2 variables. $\endgroup$ – user2520938 Feb 20 '16 at 15:22
  • $\begingroup$ How can one "plug in" $z=i$ and $\bar z=0$. That is inconsistent. $\endgroup$ – Mark Viola Feb 20 '16 at 17:36
  • $\begingroup$ @Dr.MV yes, so this functions is only defined in a subset of $\mathbb{C}^2$. I'm not looking for any kind of 'intuitive' answers. Specify the domain of $f(z,\bar{z})$, show me, if necessary by going all the way back to $\epsilon-\delta$ arguments, how we calculate $\frac{\partial f}{\partial \bar{z}}$ for this function. To start:define $f:\{(a,b)\in\mathbb{C}^2\mid a=\bar{b}\}\to \mathbb{C}:(a,b)\mapsto u\left(\frac12(a+b),\frac{1}{2i}(a-b)\right)+i v\left(\frac12(a+b),\frac{1}{2i}(a-b)\right)$ $\endgroup$ – user2520938 Feb 20 '16 at 20:26

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