Problem 10 - Advanced Engineering Mathematics

Question

(a) \(\operatorname{since} G\left(r e^{i \theta}\right)=r^{2 / 3} e^{i 2 \theta / 3}, \psi(r, \theta)=\operatorname{Im}\left(G\left(r e^{i \theta}\right)\right)=r^{2 / 3} \sin \frac{2 \theta}{3} .\) Note that \(\psi=0\) on the boundary where \(\theta=0\) and \(\theta=3 \pi / 2\) (b) \(\mathbf{V}=\overline{G^{\prime}(z)}=\overline{\frac{2}{3} z^{-1 / 3}} .\) Therefore, letting \(z=r e^{i \theta}, \mathbf{V}=\frac{2}{3} r^{-1 / 3}(\cos (\theta / 3), \sin (\theta / 3))\) for \(0<\theta<3 \pi / 2\) (c) \(r^{2 / 3} \sin (2 \theta / 3)=c\) implies that \(r=[c \csc (2 \theta / 3)]^{2 / 3}\) for \(0<\theta<3 \pi / 2\). The streamlines are shown in the figure.

Step-by-Step Solution

Answer

The function G maps the complex plane, and V characterizes flow direction.

Step 1: Understanding the Complex Function G

The function given is \( G\left(r e^{i \theta}\right) = r^{2/3} e^{i 2 \theta / 3} \). This represents a transformation in the complex plane, where the complex number \( z = r e^{i \theta} \) is mapped to another complex number after being raised to a specific power and stretched by a factor of \( r^{2/3} \), while the angle is scaled by \( 2/3 \).

Step 2: Finding the Imaginary Part

The imaginary part of the function \( G\left(r e^{i \theta}\right) \) is given by \( \operatorname{Im}(G(z)) = r^{2/3} \sin(2\theta/3) \). This represents the sine component of the complex number resulting from the transformation \( G \). It is zero when \( \theta = 0 \) and \( \theta = 3\pi/2 \), aligning with the boundaries provided.

Step 3: Evaluate Velocity Vector V

The velocity vector is given as \( \mathbf{V} = \overline{G'(z)} = \overline{\frac{2}{3} z^{-1/3}} \). By substituting \( z = r e^{i \theta} \), the velocity becomes \( \mathbf{V} = \frac{2}{3} r^{-1/3} (\cos(\theta/3), \sin(\theta/3)) \). This depicts a vector field that varies with the angle \( \theta \).

Step 4: Solve for r in terms of c and \( \theta \)

Given the equation \( r^{2/3} \sin(2\theta/3) = c \), solve for \( r \) as \( r = [c \csc(2\theta/3)]^{2/3} \). This equation characterizes the streamlines of the flow in terms of the coordinate \( \theta \). It shows how \( r \) is dependent on both the constant \( c \) and the angle.

Key Concepts

Complex Functions

Complex functions are a fascinating area of mathematics that deals with functions of a complex variable. In the context of the exercise, we are examining the function \( G\left(r e^{i \theta}\right) = r^{2/3} e^{i 2 \theta / 3} \).
This function is a type of power transformation in the complex plane. Essentially, it takes a complex number \( z = r e^{i \theta} \) and transforms it by raising its magnitude to the power of \( \frac{2}{3} \) while also scaling its angle by \( \frac{2}{3} \).

Key points to understand about this transformation include:

• It changes the size of the radius by applying a fractional power.
• The angle, or argument, of the original complex number is compressed.

By understanding these transformations, we gain insight into the geometry of complex functions and how they manipulate shapes and sizes in the complex plane.

Imaginary Part

The imaginary part of a complex function is vital because it tells us about the vertical component in the plane, derived from the sine function in polar coordinates. For the function \( G\left(r e^{i \theta}\right) \), the imaginary part is \( \operatorname{Im}(G(z)) = r^{2/3} \sin(2\theta/3) \).
When we focus on this part of the function, it expresses how much the transformed complex number moves above or below the horizontal axis.

Some important aspects to consider are:

• It takes the form of the sine of the scaled angle, which determines oscillations in the vertical direction.
• Its value becomes zero on specific boundaries, namely when \( \theta = 0 \) and \( \theta = 3\pi/2 \).

This characteristic helps us understand the function's behavior and mapping in the complex plane by focusing on how the phase changes.

Velocity Vector Field

A velocity vector field is a powerful way to visualize how complex functions change a vector space. For the given problem, the velocity vector is derived as \( \mathbf{V} = \frac{2}{3} r^{-1/3} (\cos(\theta/3), \sin(\theta/3)) \).
This expression details how the velocity is dependent on the position represented by \( r e^{i \theta} \), giving insight into flow dynamics.

Key details include:

• The vector's magnitude decreases as the radius increases, due to \( r^{-1/3} \).
• The direction of the vector is tied to a third of the original angle, illustrating rotation.

Understanding this vector field is essential for visualizing how each point in the space behaves under the influence of the complex function.

Streamlines

Streamlines represent paths followed by particles within a flow field and are crucial for visualizing complex functions' effects. In the exercise, the streamlines are described by the equation \( r^{2/3} \sin(2\theta/3) = c \), solved as \( r = [c \csc(2\theta/3)]^{2/3} \).
These streamlines provide a clear picture of how the fluid or flow behaves over time.

Important considerations are:

• They show how particles trace out certain paths, dependent on the angle \( \theta \) and a constant \( c \).
• The equation highlights the non-linear relationship between \( r \) and \( \theta \).

This visualization allows users to predict the behavior of the system and comprehend how these paths affect the overall flow dynamics, offering deeper understanding of the complex function's impact.