If f(Z) is an analytical function and (r, θ) denotes the polar co-ordinates, then:

Question

If f(Z) is an analytical function and (r, &theta;) denotes the polar co-ordinates, then:

Prepp Team · Accepted Answer

When dealing with functions in complex analysis, an analytical function, also known as a holomorphic function, is a complex-valued function that is differentiable at every point within its domain. For such a function $f(Z)$, where $Z$ is a complex variable, we can express it in terms of its real and imaginary parts. That is, $f(Z) = u(x, y) + iv(x, y)$ in Cartesian coordinates or $f(Z) = u(r, \theta) + iv(r, \theta)$ in polar coordinates.

The conditions for a complex function to be analytical are given by the Cauchy-Riemann equations. These equations relate the partial derivatives of the real part $u$ and the imaginary part $v$ of the function.

Cauchy-Riemann Equations in Cartesian Coordinates

In Cartesian coordinates, where $Z = x + iy$, the Cauchy-Riemann equations are:

$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} $
$ \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} $

Polar Coordinates Transformation

To express these equations in polar coordinates $(r, \theta)$, we use the transformation relationships between Cartesian and polar coordinates:

$ x = r \cos \theta $
$ y = r \sin \theta $

From these relationships, we can find the partial derivatives of $x$ and $y$ with respect to $r$ and $\theta$:

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

Derivation of Cauchy-Riemann Equations in Polar Form

We use the chain rule to transform the partial derivatives from Cartesian to polar coordinates. For a function $u(x, y)$ which can be expressed as $u(r, \theta)$:

$ \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} $

Substitute the partial derivatives of $x$ and $y$ with respect to $r$ and $\theta$ into these equations:

$ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta $ (Equation 1)
$ \frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x} (-r \sin \theta) + \frac{\partial u}{\partial y} (r \cos \theta) $ (Equation 2)

Similarly for $v(x, y)$ which can be expressed as $v(r, \theta)$:

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

Substitute the partial derivatives of $x$ and $y$ with respect to $r$ and $\theta$ into these equations:

$ \frac{\partial v}{\partial r} = \frac{\partial v}{\partial x} \cos \theta + \frac{\partial v}{\partial y} \sin \theta $ (Equation 3)
$ \frac{\partial v}{\partial \theta} = \frac{\partial v}{\partial x} (-r \sin \theta) + \frac{\partial v}{\partial y} (r \cos \theta) $ (Equation 4)

Deriving the First Polar Cauchy-Riemann Equation

From Equation 4, we have:

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

Now, substitute the Cartesian Cauchy-Riemann equations ($ \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} $ and $ \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} $) into the expression for $ \frac{\partial v}{\partial \theta} $:

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

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

$ \frac{\partial v}{\partial \theta} = r \left( \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta \right) $

By comparing this with Equation 1 ($ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta $), we can substitute $ \frac{\partial u}{\partial r} $ into the expression:

$ \frac{\partial v}{\partial \theta} = r \frac{\partial u}{\partial r} $

Rearranging this gives the first polar Cauchy-Riemann equation:

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

Deriving the Second Polar Cauchy-Riemann Equation

From Equation 2, we have:

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

Now, substitute the Cartesian Cauchy-Riemann equations ($ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} $ and $ \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} $) into the expression for $ \frac{\partial u}{\partial \theta} $:

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

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

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

By comparing this with Equation 3 ($ \frac{\partial v}{\partial r} = \frac{\partial v}{\partial x} \cos \theta + \frac{\partial v}{\partial y} \sin \theta $), we can substitute $ \frac{\partial v}{\partial r} $ into the expression:

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

Rearranging this gives the second polar Cauchy-Riemann equation:

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

Summary of Polar Cauchy-Riemann Equations

For an analytical function $f(Z) = u(r, \theta) + iv(r, \theta)$ in polar coordinates, the Cauchy-Riemann equations 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} $

Option Analysis

Let's compare these derived equations with the given options to find the correct match:

Option	Equations
1	$ \frac {\partial u}{\partial r} = \frac 1 r \frac {\partial v}{\partial \theta} $ and $ \frac {\partial v}{\partial r} = \frac {-1}{r} \frac {\partial u}{\partial \theta} $
2	$ \frac {\partial u}{\partial r} = \frac {-1} r \frac {\partial v}{\partial \theta} $ and $ \frac {\partial u}{\partial r} = \frac {1} r \frac {\partial v}{\partial \theta} $ (This option contains an inconsistency as it provides two different expressions for the same partial derivative $\frac{\partial u}{\partial r}$ and one of them is the same as the first part of option 1, making it likely incorrect or malformed.)
3	$ \frac {\partial u}{\partial r} = -r \frac {\partial v}{\partial \theta} $ and $ \frac {\partial v}{\partial r} = r \frac {\partial u}{\partial \theta} $
4	$ \frac {\partial u}{\partial r} = r \frac {\partial v}{\partial \theta} $ and $ \frac {\partial v}{\partial r} = -r \frac {\partial u}{\partial \theta} $

Based on our derivation, Option 1 correctly states the Cauchy-Riemann equations for an analytical function in polar coordinates.

Prepp Team · Answer

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

Prepp Team · Answer

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

Prepp Team · Answer

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

Option	Equations
1	\( \frac {\partial u}{\partial r} = \frac 1 r \frac {\partial v}{\partial \theta} \) and \( \frac {\partial v}{\partial r} = \frac {-1}{r} \frac {\partial u}{\partial \theta} \)
2	\( \frac {\partial u}{\partial r} = \frac {-1} r \frac {\partial v}{\partial \theta} \) and \( \frac {\partial u}{\partial r} = \frac {1} r \frac {\partial v}{\partial \theta} \) (This option contains an inconsistency as it provides two different expressions for the same partial derivative \(\frac{\partial u}{\partial r}\) and one of them is the same as the first part of option 1, making it likely incorrect or malformed.)
3	\( \frac {\partial u}{\partial r} = -r \frac {\partial v}{\partial \theta} \) and \( \frac {\partial v}{\partial r} = r \frac {\partial u}{\partial \theta} \)
4	\( \frac {\partial u}{\partial r} = r \frac {\partial v}{\partial \theta} \) and \( \frac {\partial v}{\partial r} = -r \frac {\partial u}{\partial \theta} \)