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My classmates and I were given that we had to verify,

\begin{eqnarray} \frac{\partial}{\partial z} (f \circ g) = (\frac{\partial f}{\partial z} \circ g)(\frac{\partial g}{\partial z}) + (\frac{\partial f}{\partial \bar{z}} \circ g)(\frac{\partial \bar{g}}{\partial z}) \end{eqnarray}

We are given the definition that $ \frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})$. This computation is quite tedious to verify and we end up with about 32 terms that needs to cancel out. For example, if $g = s(x,y) + it(x,y)$we have that the term $\frac{\partial g}{\partial z} = \frac{\partial g}{\partial x} - i \frac{\partial g}{\partial y} = \frac{\partial s}{\partial x} + i \frac{\partial t}{\partial x} + i \frac{\partial s}{\partial y} - \frac{\partial t}{\partial y}$. So, as you can see many terms are introduced very quickly.

However, our professor said this was a simple two line proof. Is there an alternative approach that will yield the verification much more quickly?

Bark
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    Try the multi-dimensional real chain rule and don't split $f,g$ into real / imaginary parts... – AlexR Sep 11 '13 at 19:06
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    What you describe sounds like a simple, two-line proof. On one line you expand into x's and y's, and on the next you cancel out everything that cancels. – Daniel McLaury Sep 11 '13 at 19:07

1 Answers1

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Just to expand a little bit on AlexR comment...

By the chain rule you may write the differential of $f\circ g$ as follows \begin{eqnarray}d(f\circ g)&=&\left(\frac{\partial f}{\partial z}\circ g \right) dg+ \left(\frac{\partial f}{\partial\bar z}\circ g \right) d\bar g\\ &=&\left(\frac{\partial f}{\partial z}\circ g \right) \left(\frac{\partial g}{\partial z}dz+\frac{\partial g}{\partial\bar z}d\bar z\right)+\left(\frac{\partial f}{\partial\bar z}\circ g \right)\left(\frac{\partial\bar g}{\partial z}dz+ \frac{\partial\bar g}{\partial\bar z}d\bar z\right)\\ &=&\left(\left(\frac{\partial f}{\partial z}\circ g \right)\frac{\partial g}{\partial z}+\left(\frac{\partial f}{\partial\bar z}\circ g\right)\frac{\partial\bar g}{\partial z}\right)dz+ \left(\left(\frac{\partial f}{\partial z}\circ g \right)\frac{\partial g}{\partial\bar z}+\left(\frac{\partial f}{\partial\bar z}\circ g\right)\frac{\partial\bar g}{\partial\bar z}\right)d\bar z \end{eqnarray} Identifying the "coefficients" in front of $dz$ and $d\bar z$, this gives you both $\frac{\partial(f\circ g)}{\partial z} $ and $\frac{\partial(f\circ g)}{\partial\bar z} \cdot$

mucciolo
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Etienne
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  • Thank you very much for expanding on that comment! However, I do not understand where you are getting the second term involving $d \bar{g}$. Could you explain where that is coming from? – Bark Sep 11 '13 at 19:38
  • Start with $df=\frac{\partial f}{\partial z} dz+\frac{\partial f}{\partial\bar z}d\bar z$. The chain rule says exactly that for computing $d(f\circ g)$, you just have to replace the variable $z$ with the function $g$, i.e. to replace $dz$ with $dg$ and $d\bar z$ with $d\bar g$. – Etienne Sep 11 '13 at 19:46
  • Ahh, I see! Thank you so much! This is far more elegant than what I was working with! – Bark Sep 11 '13 at 19:49
  • Sorry but I am still stuck in you notation of $dg,dz$ what does that mean ? – user162343 Feb 10 '16 at 17:48
  • Don't we need to replace $\partial z$ by $\partial f$ inside the parentesis?? – bttmbrcelo Feb 01 '17 at 13:58