Question

Let \(\mathbb{R}\) denote the set of all real numbers and let \(i = \sqrt{-1}\). Consider the matrices \[ S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \quad \text{and} \quad T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. \] Let \(a, b, c, d\) be real numbers such that \[ ST = \begin{pmatrix} a & b \\ c & d \end{pmatrix}. \] Let \[ H = \{x + iy : x, y \in \mathbb{R} \text{ and } y>0\}. \] Then which of the following statements is (are) TRUE?

• A. $\frac{b + ia}{d + ic} = i$
• B. If $\omega = \frac{-1 + i\sqrt{3}}{2}$, then $\frac{a\omega + b}{c\omega + d} = \omega$
• C. If $m$ is an integer greater than 2 such that $(ST)^2 = (ST)^m$, then $m$ is an integer multiple of 8
• D. If $z \in H$, then $\frac{az + b}{cz + d} \in H$

Hint:

Transformations of the form $f(z) = \frac{az+b}{cz+d}$ with $ad-bc > 0$ and real coefficients map the upper half-plane to itself. In this case, although $a=0$, the property holds because the determinant of the matrix $ST$ is 1 (positive).

Answer 1

The Correct Option is B

To determine which of the given statements is true, let's start by finding the product of the matrices \( S \) and \( T \).

\( S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \)

\( T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \)

The matrix multiplication of \( S \) and \( T \) is calculated as follows:

\( ST = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} (0 \cdot 1 + (-1) \cdot 0) & (0 \cdot 1 + (-1) \cdot 1) \\ (1 \cdot 1 + 0 \cdot 0) & (1 \cdot 1 + 0 \cdot 1) \end{pmatrix} \)

\( = \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix} \)

So we have the matrix \( ST = \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix} \) implying \( a = 0 \), \( b = -1 \), \( c = 1 \), and \( d = 1 \).

1. To check the statement \(\frac{b + ia}{d + ic} = i\):
   • Substitute the values: \(\frac{-1 + i \cdot 0}{1 + i \cdot 1} = \frac{-1}{1 + i}\)
   • Rationalizing the denominator: \(\frac{-1(1 - i)}{(1 + i)(1 - i)} = \frac{-1 + i}{1^2 - i^2} = \frac{-1 + i}{2}\)
   • Clearly, \(\frac{-1}{2} + \frac{i}{2} \neq i\), so this statement is false.
2. To check the statement \(\frac{a\omega + b}{c\omega + d} = \omega\):
   • Given that \(\omega = \frac{-1 + i\sqrt{3}}{2}\)
   • Calculate the expression: \(\frac{0 \cdot \omega - 1}{1 \cdot \omega + 1} = \frac{-1}{\omega + 1}\)
   • Simplify \(\omega+1 = \frac{-1 + i\sqrt{3}}{2} + 1 = \frac{1 + i\sqrt{3}}{2}\)
   • Now, \(\frac{-1}{\omega + 1} = \frac{-2}{1 + i\sqrt{3}} \cdot \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}} = \frac{-2 + 2i\sqrt{3}}{1 + 3} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1}{2} + \frac{i\sqrt{3}}{2} = \omega\)
   • This statement is true.
3. To check whether \( m \) is an integer multiple of 8:
   • Given \((ST)^2 = (ST)^m \Rightarrow (ST)^{m-2} = I\), where \( I \) is the identity matrix.
   • The order of \(ST\) is determined by calculating successive powers, which repeats every 4 steps. Therefore, \(m-2\) must be a multiple of 4.
   • Thus, \(m\) itself must be a multiple of 8 for the equation to hold, making this statement true.
4. To verify if \(\frac{az + b}{cz + d} \in H\):
   • If \(z \in H = \{x + iy : x, y \in \mathbb{R}, y > 0\}\), then \(\text{Im}\left(\frac{az + b}{cz + d}\right) > 0\) needs to be verified, which holds for all \( z \) in upper half plane since \(|ST| = 1\).
   • This statement is also true.

The correct options would be the second and third.