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This is a question from Stein and Shakarchi's book 'Complex Analysis', chapter 1, exercise 8, page 27.

We have two differentiable functions $f:U \to V$ and $g:V \to \mathbb{C}$ where $U$ and $V$ are two open subsets of $\mathbb{C}$. Let $h=g \circ f$ and define the differential operators

$$ \frac{\partial}{\partial z} =\frac{1}{2}\left(\frac{\partial}{\partial x} +\frac{1}{i}\frac{\partial}{\partial y} \right) $$ and $$ \frac{\partial}{\partial \bar{z}} =\frac{1}{2}\left(\frac{\partial}{\partial x} -\frac{1}{i}\frac{\partial}{\partial y}\right) $$ as in page 12.

The first part asks us to show that

$$ \frac{\partial h}{\partial z} =\left(\frac{\partial g}{\partial z} \circ f\right) \, \frac{\partial f}{\partial z} +\left(\frac{\partial g}{\partial {\bar{z}}} \circ f \right) \, \frac{\partial \bar{f}}{\partial{z}} $$.

I am wondering if there is a mistake in the question because $$\frac{\partial g}{\partial {\bar{z}}}=0$$ by the Cauchy-Riemann equations as g is differentiable, and so the second product is zero. But I do obtain the first product using the usual chain rule for two real variables.

This question has been asked before on SE: The complex version of the chain rule but I do not understand the answer given there, which introduces the variable $w=f(z,\bar{z})$ and then assumes that $w$ and $\bar{w}$ are independent.

I'd appreciate your help!

Joe Y
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    The page you linked to also has a link in it. Did you see it? It might help. https://math.stackexchange.com/questions/490863/homework-problem-complex-analysis-chain-rule?rq=1 –  Nov 25 '19 at 20:08
  • Many thanks for that but I still want to prove the formula using the chain rule for two real variables using the definition of $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$. I've tried many times but every time I only get the first two terms. – Joe Y Nov 25 '19 at 22:47
  • I like the answer linked by tilper because partial derivatives are by right the coefficients of the differential. In the answer OP links, Mark Viola uses (equivalent) coordinates $z, \overline{z}$ instead of $x, y$ (which is why one introduces Wirtinger derivatives). If you wish to stick to real coordinates, you should aim to Debussy formulae (in the same thread as Viola) and then conclude. Of course the formula holds for complex-valued differentiable functions; if your functions are holomorphic, ie satisfy the delta bar equation $\overline{\partial}g = 0$, then you get the usual chain rule. –  Nov 26 '19 at 11:55
  • For the independence: if ${x, iy}$ are independent then so are${z, \overline{z}}$, as those coordinates are connected by a linear isomorphism (the Jacobian is an invertible matrix, and moreover is constant). –  Nov 26 '19 at 12:02
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    @piombino Many thanks for your comments, they are very helpful. I want to vote them up but unfortunately I can't as I'm having some problems with my browser. – Joe Y Nov 26 '19 at 22:55

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