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For some reason I have trouble recognising why $$\frac{\partial z}{\partial \bar{z}}=0$$ which is used countlessly in complex analysis, because it appears to me there should be some dependence between $z$ and $\bar{z}$. Why can they be treated an independent variables?

Zion R.
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    try to write $z$ as $f(\bar{z})$ and make the derivate. For example, you can use that $Re(z) = (z+\bar{z})/2$ to find a relation between $z$ and $\bar{z}$. Or you can write the full form of the conjugations transformation – luisfelipe18 Oct 27 '21 at 17:15
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    Certainly there is "some dependence" between $z$ and $\bar z$, and the one is not a constant function of the other. If you tell me $\bar z$, then I can tell you $z$. But this functional dependence is not a differentiable function in the context of a function of a single complex variable. Perhaps you can identify a complex analysis textbook where that "is used countlessly"? – hardmath Oct 27 '21 at 17:21
  • @hardmath For example, when $x=\frac{1}{2}(z+\bar{z})$, you can partially differentiate $x$ by $\bar{z}$ and get $\frac{1}{2}$, which appears to treat $z$ as independent of $\bar{z}$. – Zion R. Oct 27 '21 at 17:27
  • If your point is that it doesn't make sense to "treat $z$ as independent of $\bar z$" (in the sense that complex number $z$ cannot vary without also having its conjugate $\bar z$ vary), then I agree with you. But your original post was not about partial differentiation, so I'm still hoping you can cite a textbook that uses the formula you wrote there. – hardmath Oct 27 '21 at 17:41
  • @hardmath The course I did had its own lecture notes so isn't available online. When I say countlessly, I more mean that results that appear very often appear to require the partial derivative I wrote to equal zero. I checked in Chaum's book just now and saw that the del operator is rather defined to be $\nabla=\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}=2\frac{\partial}{\partial \bar{z}}$, and the fact in question follows quickly from this definition, when using the example I gave in my previous comment. (What do you mean that my original post was not about partial diff? It was) – Zion R. Oct 27 '21 at 18:39
  • It makes sense to treat $z = x+iy$ as a function of two real variables, so you can find partial derivatives of $z$ with respect to $x,y$. But $z$ is uniquely determined by $\bar z$, so if you want to have a partial derivative, you need more than one argument (and that the arguments can vary independently). I believe you are trying to ask something about that, but your very brief Question post does not tell Readers what variables are to be treated as independent arguments. – hardmath Oct 27 '21 at 18:50

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