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## Problem

Consider the complex number in polar form $z = r{e}^{i \theta}$, and the complex function in polar form $f(z) = u(r, \theta) + iv(r, \theta)$.

Show that the Cauchy-Riemann equations in polar form are

$\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}$
$\frac{\partial v}{\partial r} = -\frac{1}{r} \frac{\partial u}{\partial \theta}$.

## Solution

It is important to know the Cartesian representation of the complex number and the Cauchy-Riemann equations first.

$z = x+iy$
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$
$\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$.

Now use the multivariate product rule to take the the derivatives of $\frac{\partial u}{\partial r}$, $\frac{\partial u}{\partial \theta}$, $\frac{\partial v}{\partial r}$, and $\frac{\partial v}{\partial \theta}$.

For the function $u(r, \theta)$:

$\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r}$
$\frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta}$.

Since $\frac{\partial x}{\partial r} = \cos{\theta}$, $\frac{\partial y}{\partial r} = \sin{\theta}$, $\frac{\partial x}{\partial \theta} = -r\sin{\theta}$, and $\frac{\partial y}{\partial \theta} = r\cos{\theta}$, we get

$\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos{\theta} + \frac{\partial u}{\partial y} \sin{\theta}$
$\frac{\partial u}{\partial \theta} = -\frac{\partial u}{\partial x} r\sin{\theta} + \frac{\partial u}{\partial y} r\cos{\theta}$.

We repeat the same calculations for $v(r, \theta)$.

$\frac{\partial v}{\partial r} = \frac{\partial v}{\partial x} \cos{\theta} + \frac{\partial v}{\partial y} \sin{\theta}$
$\frac{\partial v}{\partial \theta} = -\frac{\partial v}{\partial x} r\sin{\theta} + \frac{\partial v}{\partial y} r\cos{\theta}$.

Now use the Cauchy-Riemann equations in Cartesian form and make the right substitutions.

$\frac{\partial v}{\partial \theta} = \frac{\partial u}{\partial y} r\sin{\theta} + \frac{\partial u}{\partial x} r\cos{\theta} = r\frac{\partial u}{\partial r}$
$\frac{\partial u}{\partial \theta} = -\frac{\partial v}{\partial y} r\sin{\theta} - \frac{\partial v}{\partial x} r\cos{\theta} = -r\frac{\partial v}{\partial r}$

Therefore,

$\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}$
$\frac{\partial v}{\partial r} = -\frac{1}{r} \frac{\partial u}{\partial \theta}$.
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