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

Step 1: Understanding the Question:
The problem explores matrix products, Mobius transformations, and matrix periodicity.

Step 2: Key Formula or Approach:

• Compute $ST$.

• Evaluate the fractional transformation for $\omega$ and a general $z$ in the upper half-plane.

• Use the characteristic equation or powers of $ST$ to find periodicity.

Step 3: Detailed Explanation:

• $ST = \begin{pmatrix} 0 & -1
1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1
0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1
1 & 1 \end{pmatrix}$. Thus $a=0, b=-1, c=1, d=1$.

• Checking (A): $\frac{b+ia}{d+ic} = \frac{-1 + i(0)}{1 + i(1)} = \frac{-1}{1+i} = \frac{-1(1-i)}{2} = \frac{-1+i}{2}$. Statement (A) is FALSE.

• Checking (B): $\frac{a\omega + b}{c\omega + d} = \frac{0\omega - 1}{1\omega + 1} = \frac{-1}{\omega + 1}$. Since $\omega^2 + \omega + 1 = 0 \implies \omega + 1 = -\omega^2$: \[ \frac{-1}{-\omega^2} = \frac{1}{\omega^2} = \frac{\omega^3}{\omega^2} = \omega. \] Statement (B) is TRUE.

• Checking (C): $M = ST = \begin{pmatrix} 0 & -1
1 & 1 \end{pmatrix}$. Characteristic equation: $\lambda^2 - \lambda + 1 = 0$. Roots are $e^{i\pi/3}, e^{-i\pi/3}$. This means $M^6 = I$. $M^2 = M^m \implies M^{m-2} = I \implies m-2 = 6k \implies m = 6k + 2$. For $k=1$, $m=8$ (multiple of 8). For $k=2$, $m=14$ (not a multiple of 8). Statement (C) is FALSE.

• Checking (D): For $z \in H$, $\text{Im}(z) > 0$. Transformation is $f(z) = \frac{-1}{z+1}$. \[ \text{Im}(f(z)) = \text{Im}\left( \frac{-1}{x+1+iy} \right) = \text{Im}\left( \frac{-(x+1-iy)}{(x+1)^2+y^2} \right) = \frac{y}{(x+1)^2+y^2}. \] Since $y > 0$, $\text{Im}(f(z)) > 0$. So $f(z) \in H$. Statement (D) is TRUE.

Step 4: Final Answer:
The true statements are (B) and (D).