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I already found $\partial_z \bar{z}=\lim_{h \rightarrow 0} \frac{\bar{h}}{h}$, but I'm confused about

$$\partial_z \frac{1}{\bar{z}}$$ I got

$$\partial_z \frac{1}{\bar{z}}=\lim_{h \rightarrow 0} \frac{\frac{1}{\over{z+h}}-\frac{1}{\bar{z}}}{h}=\lim_{h \rightarrow 0}\frac{1}{\over{z(z+h)}}\frac{\bar{h}}{h}$$

But I cannot see what to do with this.

Can I really conclude that it doesn't exist just because $\frac{\bar{h}}{h}$ doesn't?

I can also do:

$$\partial_z \frac{1}{\bar{z}}=\lim_{h \rightarrow 0} \frac{1}{h}({\frac{1}{\over{z+h}}-\frac{1}{\bar{z}}})$$

But what's this?

mavavilj
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  • Or should I be using Wirtinger derivative definition? – mavavilj Feb 16 '16 at 12:25
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    The Wirtinger derivatives are in general not limits of difference quotients. To compute $\partial_z (1/\overline{z})$, you need to work differently. Which definition(s) of the Wirtinger derivatives are you working with? – Daniel Fischer Feb 16 '16 at 12:26
  • @DanielFischer This one: https://en.wikipedia.org/w/index.php?title=Wirtinger_derivatives§ion=5#Functions_of_one_complex_variable – mavavilj Feb 16 '16 at 12:28
  • Okay. I guess classical derivative on $\mathbb{C}\setminus {0}$ rather than distributional derivative on $\mathbb{C}$. In either case, it may be helpful to recall/observe that $\partial_z f = \overline{\partial_{\overline{z}} \overline{f}}$. – Daniel Fischer Feb 16 '16 at 12:36
  • @DanielFischer What's $\partial_{\bar{z}}$ if it's not the Wirtinger derivative definition? – mavavilj Feb 16 '16 at 12:41
  • It's the Wirtinger derivative. The point is that you probably already know what $\partial_{\overline{z}} \frac{1}{z}$ is, so that saves you some computation. – Daniel Fischer Feb 16 '16 at 12:43
  • @DanielFischer But I computed $\partial_z \bar{z}$ using the limit definition and not Wirtinger. Are these interchangeable? – mavavilj Feb 16 '16 at 12:48
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    Actually, you didn't. First, there is no limit definition for the Wirtinger derivatives. In some circumstances, you can get the value of the Wirtinger derivative[s] as a limit of difference quotients, but generally you can't. For $f(z) = \overline{z}$, the difference quotient $\frac{f(z+h) - f(z)}{h}$ evaluates to $\overline{h}/h$, but the limit of that as $h\to 0$ doesn't exist. Second, in your "$\partial_z \overline{z} = \frac{\overline{h}}{h}$", the right hand side depends on $h$, but the left doesn't. – Daniel Fischer Feb 16 '16 at 12:55
  • @DanielFischer I meant that I used the limit definition and not Wirtinger. Should the Wirtinger produce the same result however? – mavavilj Feb 16 '16 at 12:56
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    The limit definition of what? We have a limit definition of the complex derivative, but that applies only to functions that are complex differentiable at the point in question. Well, a function $f$ is complex differentiable at $z$ if and only if $\lim\limits_{h\to 0} \frac{f(z+h) - f(z)}{h}$ exists. If you look at $f(z) = \overline{z}$ or $f(z) = 1/\overline{z}$, and form the difference quotients, you will see that that limit doesn't exist for any $z$ in the domain of $f$. But these functions are (infinitely often) real-differentiable, so the Wirtinger derivatives exist at all $z$. – Daniel Fischer Feb 16 '16 at 13:03

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As Daniel Fischer pointed out, it makes no sense to compute $\partial_z\frac{1}{\overline{z}}$ by limits of difference quotients, since $\frac{1}{\overline{z}}$ is not complex differentiable.

Therefore, you can only use the Wirtinger definition of $$\partial_z := \frac{1}{2}(\partial_x-i\partial_y).$$ By computing those partial derivatives you should obtain $$\partial_z\frac{1}{\overline{z}}=\frac{1}{2}\frac{-1-i^2}{(x-iy)^2}=0,$$ which is no surprise, since you might also observe $$\partial_z\frac{1}{\overline{z}}=\overline{\partial_\overline{z}\frac{1}{z}},$$ where the latter is $0$ since $\partial_\overline{z}$ is $0$ for any complex differentiable function.

TheAbelian
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