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How do i write $\frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial \bar z}$ in polar form.

From my textbook I know, $$\frac{\partial f}{\partial z} = \frac{e^{-i\theta}}{2} \left(\frac{\partial }{\partial r} - \frac {i}{r} \frac{\partial }{\partial \theta} \right)$$

$$\frac{\partial f}{\partial \bar z} = \frac{e^{i\theta}}{2} \left(\frac{\partial }{\partial r} + \frac {i}{r} \frac{\partial }{\partial \theta} \right)$$

how do i derive these answers?

i know $$ z = re^{i\theta}$$

and, $$ z = u(r,\theta) + iv(r,\theta)$$

HAC
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1 Answers1

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Start with $z=re^{i\theta}$ (i.e. $\bar{z}=re^{-i\theta}$) and apply the chain rule

$$\begin{align} {\partial\over\partial r}&={\partial\over\partial z}{\partial z\over\partial r}+{\partial\over\partial \bar{z}}{\partial \bar{z}\over\partial r}\\ {\partial\over\partial \theta}&={\partial\over\partial z}{\partial z\over\partial \theta}+{\partial\over\partial \bar{z}}{\partial \bar{z}\over\partial \theta} \end{align}$$

One has

$$\begin{align} {\partial z\over\partial r}&=e^{i\theta}\\ {\partial \bar{z}\over\partial r}&=e^{-i\theta}\\ {\partial z\over\partial\theta}&=ire^{i\theta}\\ {\partial\bar{z}\over\partial\theta}&=-ire^{-i\theta} \end{align}$$

Now substitute

$$\begin{align} {\partial\over\partial r}&={\partial\over\partial z}e^{i\theta}+{\partial\over\partial \bar{z}}e^{-i\theta}\\ {\partial\over\partial \theta}&=ir {\partial\over\partial z}e^{i\theta}-ir {\partial\over\partial \bar{z}}e^{-i\theta} \end{align}$$

Now inverse the system to get the expected result.

marwalix
  • 16,773
  • what do you mean by inversing the system? – HAC Jan 15 '18 at 13:31
  • we have a linear system of two equations in the “unknown” $\partial/\partial z$ and $\partial/\partial\bar{z}$. I will add the details in the answer as soon as I am back behind my desktop – marwalix Jan 15 '18 at 16:36
  • thanks. i was too caught up in the differentials tht i didnt realize the system of equations. i got it now. – HAC Jan 15 '18 at 17:59