Problem 8 - Advanced Engineering Mathematics

Question

Use the matrix method to compute \(S^{-1}(w)\) and \(S^{-1}(T(z))\) for each pair of linear fractional transformations. \(T(z)=\frac{z-1+i}{i z-2} \quad\) and \(\quad S(z)=\frac{(2-i) z}{z-1-i}\)

Step-by-Step Solution

Answer

Use the inverse transformation formula to find \(S^{-1}(z)\) and combine with \(T(z)\) for \(S^{-1}(T(z))\).

Step 1: Understand the Linear Fractional Transformation

Linear fractional transformations are mappings of the form \(f(z) = \frac{az + b}{cz + d}\). For the transformations given: \(T(z) = \frac{z - 1 + i}{iz - 2}\) and \(S(z) = \frac{(2-i)z}{z-1-i}\), identify \(T(z)\) with coefficients \(a=1, b=-1+i, c=i, d=-2\) and \(S(z)\) with \(a=2-i, b=0, c=1, d=-(1+i)\).

Step 2: Construct Corresponding Matrices

For a linear fractional transformation \(f(z) = \frac{az + b}{cz + d}\), associate it with the matrix \(M = \begin{pmatrix} a & b \ c & d \end{pmatrix}\). Thus, for \(T(z)\), the matrix is \(M_T = \begin{pmatrix} 1 & -1+i \ i & -2 \end{pmatrix}\), and for \(S(z)\), the matrix is \(M_S = \begin{pmatrix} 2-i & 0 \ 1 & -(1+i) \end{pmatrix}\).

Step 3: Find the Inverse of the Matrix

To find the inverse \(M^{-1}\) of matrix \(M = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), use the formula \(M^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \ -c & a \end{pmatrix}\). First, compute the determinant for each matrix: - For \(M_S\), \(ad-bc = (2-i)(-(1+i)) - (0) = -3 + i\). - For \(M_T\), \(ad-bc = (1)(-2) - (-1+i)(i) = -3 - i\).

Step 4: Calculate \(S^{-1}(z)\) and \(T^{-1}(z)\)

Now use the formula to compute inverse matrices:- For \(M_S\), \[ M_S^{-1} = \frac{1}{-3+i} \begin{pmatrix} -(1+i) & 0 \ -1 & (2-i) \end{pmatrix} = \begin{pmatrix} \frac{1+i}{-3+i} & 0 \ \frac{1}{-3+i} & \frac{2-i}{-3+i} \end{pmatrix} \]- For \(M_T\), \[ M_T^{-1} = \frac{1}{-3-i} \begin{pmatrix} -2 & 1-i \ -i & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{-3-i} & \frac{1-i}{-3-i} \ \frac{i}{-3-i} & \frac{1}{-3-i} \end{pmatrix} \].

Step 5: Compose Transformations to Find \(S^{-1}(T(z))\)

To find \(S^{-1}(T(z))\), multiply the inverse of \(S(z)\) with \(T(z)\) using matrix multiplication: - \(S^{-1}(x)\) corresponds to \(M_S^{-1}\cdot M = \begin{pmatrix} \frac{1+i}{-3+i} & 0 \ \frac{1}{-3+i} & \frac{2-i}{-3+i} \end{pmatrix} \begin{pmatrix} 1 & -1+i \ i & -2 \end{pmatrix}\). Perform the multiplication to calculate the resulting matrices for \(S^{-1}(w)\) and \(S^{-1}(T(z))\).

Step 6: Calculate and Simplify the Resulting Expressions

Compute the products and simplify to obtain the transformation functions to derive the expressions for \(S^{-1}(w)\) and \(S^{-1}(T(z))\). This involves simplifying the complex number operations based on the transformations computed from earlier steps.

Key Concepts

Matrix Method

Linear fractional transformations can be approached using matrices, which makes their manipulation more structured. This method associates a specific matrix to each transformation function. Consider the transformation function represented by \( f(z) = \frac{az + b}{cz + d} \). The corresponding matrix can be written as \( M = \begin{pmatrix} a & b \ c & d \end{pmatrix} \).

For example:

• Transformation \( T(z) = \frac{z - 1 + i}{iz - 2} \): Its matrix is \( M_T = \begin{pmatrix} 1 & -1+i \ i & -2 \end{pmatrix} \).
• Transformation \( S(z) = \frac{(2-i)z}{z-1-i} \): Its matrix is \( M_S = \begin{pmatrix} 2-i & 0 \ 1 & -(1+i) \end{pmatrix} \).

Using matrices in this way provides a neat form for calculations like finding inverses and composing transformations.

Inverse Transformation

To reverse the effect of a linear fractional transformation, we find its inverse transformation, often achieved via the inverse of its matrix. Given a matrix \( M = \begin{pmatrix} a & b \ c & d \end{pmatrix} \), its inverse \( M^{-1} \) can be calculated using:
\[ M^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \ -c & a \end{pmatrix} \]

Here's how it's done:

• For \( M_S \), with coefficients \( a=2-i, b=0, c=1, d=-(1+i) \), the determinant is \( ad-bc = -3+i \). The inverse, \( M_S^{-1} \), is \( \begin{pmatrix} \frac{1+i}{-3+i} & 0 \ \frac{1}{-3+i} & \frac{2-i}{-3+i} \end{pmatrix} \).
• For \( M_T \), with \( a=1, b=-1+i, c=i, d=-2 \), the determinant is \( -3-i \). The inverse, \( M_T^{-1} \), becomes \( \begin{pmatrix} \frac{2}{-3-i} & \frac{1-i}{-3-i} \ \frac{i}{-3-i} & \frac{1}{-3-i} \end{pmatrix} \).

Finding the inverse is crucial for operations such as solving \( S^{-1}(T(z)) \).

Complex Analysis

In complex analysis, linear fractional transformations are fundamental as they represent bijective conformal mappings of the extended complex plane. These transformations can be viewed as compositions of simpler operations such as translation, rotation, and scaling.

Let's dig into the transformations:

• Translation in the complex plane shifts the entire plane by a constant vector.
• Rotation involves multiplying the complex number by a unit magnitude complex number.
• Scaling changes the size of the complex number plane but keeps its structure.

The transformation \( T(z) = \frac{z - 1 + i}{iz - 2} \) and \( S(z) = \frac{(2-i)z}{z-1-i} \), through matrices, can alter the role of these operations fluidly. Multiplying entire transformations using matrices reflects combining these operations to create new mappings.