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In my text on complex analysis, they give the definition of $\frac{\partial f}{\partial \overline{z}}$ for suitable $f : \mathbb{C} \rightarrow \mathbb{C}$. However, I do not understand how to make formal the fact that if you write $f$ 'in terms of' $z$ and $\overline{z}$, $\frac{\partial f}{\partial \overline{z}}$ is just the partial derivative with respect to $\overline{z}$ of $f$, as if you substituted $x=z$ and $y=\overline{z}$ and then took the partial with respect to $y$. The book uses this without proof, I would like to understand why this works. Thanks in advance.

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The reason your book uses the expressions $\frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial \overline{z}}$ without proof is that they are not actual partial derivatives, they are only derivatives in a symbolic sense. They are formal expressions that can help in understanding and communicating the behavior of a complex function.

If we write $x = \frac12(z + \overline{z})$ and $y = \frac{1}{2i}(z - \overline{z})$, thus expressing $x$ and $y$ as functions of $z$ and $\overline{z}$, and think of $f(z) = f(x,y)$ as a function of two real variables, then by supposing the rules of calculus apply we obtain

$$ \frac{\partial f}{\partial z} = \frac 12 \left(\frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y} \right) $$

and

$$ \frac{\partial f}{\partial \overline{z}} = \frac 12 \left(\frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right). $$

If $\frac{\partial f}{\partial \overline{z}} = 0$ for a particular function, then this is equivalent with saying that $f$ satisfies the Cauchy-Riemann equations. So in a formal sense we could say that an analytic function is a function of $z$ alone and independent of $\overline{z}$, thinking of them independent variables. (Ahlfors)

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    $\begingroup$ In my book they for example say $$ \frac{\partial}{\partial \overline{z}}\overline{z}^3=3\overline{z}^2$$, not by using the formula you give(which is also in my text), but by saying that the partial with respect to $\overline{z}$ just does what you would expect it does by its name. I would like to understand how this follows from the definition for any arbitrary function. $\endgroup$ – M. Van May 21 '16 at 19:11
  • $\begingroup$ By giving that example and remarking that the partial derivative with respect to $\overline{z}$ 'does what you would expect it does by its name,' they are just emphasizing that even though it is only a derivative in a symbolic sense, not in terms of limits, it behaves just like you would expect if it was a real derivative. They just did not show the details, but the computation is by using the definition I gave above for $\frac{\partial f}{\partial {\overline{z}}}$. $\endgroup$ – user333870 May 21 '16 at 19:20

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