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What am I doing wrong?

Let $\omega =f(z)=R(r,\theta)e^{i \,\Theta(r,\theta)} $

I want to find Cauchy Riemann conditions by differentiating in respect to $\theta $ and r and then equalizing the results, i.e.: $$ \frac{\partial}{\partial r}(Re^{i \,\Theta})=\frac{\partial}{\partial (i\theta)}(Re^{i \,\Theta})$$

where $R \,and \,\Theta$ are functions of $r \,and \,\theta$

From the left side of the equation I get (applying the product and chain rule):

$\frac{\partial}{\partial r}(Re^{i \,\Theta})= R ie^{i\Theta}\frac{\partial \Theta}{\partial r}+e^{i\Theta}\frac{\partial R}{\partial r}$

$\frac{\partial}{\partial (i\theta)}(Re^{i \,\Theta})= (-i \cdot i) R e^{i\Theta}\frac{\partial \Theta}{\partial \theta}-i\cdot e^{i\Theta}\frac{\partial R}{\partial \theta} = R e^{i\Theta}\frac{\partial \Theta}{\partial \theta}- i\cdot e^{i\Theta}\frac{\partial R}{\partial \theta}$

If we equalize the real an imaginary parts then we have: $$ \frac{\partial R}{\partial r} = R \frac{\partial}{\partial \theta}$$ and $$ R\frac{\partial \Theta}{\partial r} = -\frac{\partial R}{\partial \theta}$$

However, in my book (Arfken) the Cauchy Riemann relation in polar coordinates are: $$ \frac{\partial R}{\partial r} = \frac{R}{r} \frac{\partial}{\partial \theta}$$ and $$-R\frac{\partial \Theta}{\partial r} = \frac{1}{r}\frac{\partial R}{\partial \theta}$$

Why am I missing a factor of $\frac{1}{r}$?

Ivan
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  • This might help: Cauchy – Hushus46 Aug 24 '17 at 20:32
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    The equation you start with, $\frac{\partial}{\partial r}(Re^{i ,\Theta})=\frac{\partial}{\partial (i\theta)}(Re^{i ,\Theta})$, is wrong. Instead you should require $i \frac{\partial}{\partial r}(Re^{i ,\Theta})=\frac{1}{r} \frac{\partial}{\partial \theta}(Re^{i ,\Theta})$. You can try testing these on a simple function like $f(z) = z = re^{i \theta}$ to see which one works (and then look for the explanation). – Michał Miśkiewicz Aug 24 '17 at 21:01
  • https://math.stackexchange.com/questions/1245754/cauchy-riemann-equations-in-polar-form – Nosrati Aug 25 '17 at 07:51

1 Answers1

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The derivative in the direction of $\theta$ is actually wrong

The derivative in that direction should be: $$ \frac{\partial R e^{i \Theta}}{\partial (ri d\theta)} = \frac{\partial R e^{i \Theta}}{ri \,\partial \theta}$$

This because the length differential isn't only $\partial \theta$ but $r\partial \theta$

This because the arc is longer when we move the same $d \theta$ in a larger distance (radius) from the origin

So the derivatives that should be equalized are:

$$\frac{\partial R e^{i \Theta}}{ \,\partial r} = \frac{1}{r}\frac{\partial Re^{i\Theta}}{i\partial \theta}$$

This explains the factor $\frac{1}{r}$

Ivan
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