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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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2 Answers2

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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
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  • 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
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So $z=re^{i\theta}$ and $f=u(r,\theta)+iv(r,\theta)$

Now $z+\Delta z=(r+\Delta r)e^{i(\theta + \Delta\theta)}$ $$\Rightarrow \Delta z=e^{i\theta}(re^{i\Delta\theta}+\Delta re^{i\Delta\theta}-r)$$

Using Euler's formula and approximating $\cos\Delta\theta\to1$ and $\sin\Delta\theta\to\Delta\theta$,

$$\Delta z=e^{i\theta}(r+ir\Delta\theta+\Delta r+i\Delta r\Delta\theta-r)$$

Here, the term $\Delta r\Delta\theta$ is negligible. So, $$\Delta z=e^{i\theta}(\Delta r+ir\Delta\theta)$$

By the definition of a derivative,

$${df\over dz}=\lim_{\Delta z\to0}{f(z+\Delta z)-f(z) \over\Delta z}$$

$${df\over dz}=\lim_{\Delta r\to0,\Delta\theta\to0}{f(r+\Delta r,\theta+\Delta\theta)-f(r,\theta) \over e^{i\theta}(\Delta r+ir\Delta\theta)}$$

If we let $\Delta\theta\to0$ first,

$${df\over dz}=e^{-i\theta}\lim_{\Delta r\to0}{f(r+\Delta r,\theta)-f(r,\theta) \over\Delta r}$$ $${df\over dz}=e^{-i\theta}{\partial f\over\partial r}\tag1$$

But if we let $\Delta r\to0$ first, $${df\over dz}=e^{-i\theta}\lim_{\Delta\theta\to0}{f(r,\theta+\Delta\theta)-f(r,\theta) \over ir\Delta\theta}$$ $${df\over dz}={-i\over r}e^{-i\theta}{\partial f\over\partial\theta}\tag2$$

Adding equations 1 and 2 and dividing by 2 gives, $${df\over dz}={e^{-i\theta}\over2}\left({\partial f\over\partial r}-{i\over r}{\partial f\over\partial\theta}\right)\tag{QED.}$$

Use the same reasoning to derive ${df\over d\bar z}$

Edward Saveri
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