Let ℝ denote the set of all real numbers and let i = -1 . Consider...

We are given the matrices $S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ and $T = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.

First, we compute the product $ST$:
$ST = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 1 \end{bmatrix}$

Comparing this with $ST = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, we get $a = 0$, $b = -1$, $c = 1$, and $d = 1$.

Evaluating Option (A):
$\dfrac{b + ia}{d + ic} = \dfrac{-1 + i(0)}{1 + i(1)} = \dfrac{-1}{1 + i} = \dfrac{-1(1 - i)}{(1 + i)(1 - i)} = \dfrac{-1 + i}{2}$
This is not equal to $i$, so statement (A) is FALSE.

Evaluating Option (B):
Given $\omega = \dfrac{-1 + i\sqrt{3}}{2}$, which is a complex cube root of unity, satisfying $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$.
$\dfrac{a\omega + b}{c\omega + d} = \dfrac{0 \cdot \omega - 1}{1 \cdot \omega + 1} = \dfrac{-1}{\omega + 1}$
Since $\omega + 1 = -\omega^2$, we have:
$\dfrac{-1}{-\omega^2} = \dfrac{1}{\omega^2} = \dfrac{\omega^3}{\omega^2} = \omega$
Thus, statement (B) is TRUE.

Evaluating Option (C):
The characteristic equation of $ST$ is $\lambda^2 - \text{Tr}(ST)\lambda + \det(ST) = 0$, which gives $\lambda^2 - \lambda + 1 = 0$.
By the Cayley-Hamilton theorem, $(ST)^2 - ST + I = 0$.
Multiplying by $(ST + I)$, we get $(ST)^3 + I = 0 \Rightarrow (ST)^3 = -I$.
Squaring both sides gives $(ST)^6 = I$.
We are given $(ST)^2 = (ST)^m$, which implies $(ST)^{m-2} = I$.
This means $m - 2$ must be a multiple of $6$, so $m = 6k + 2$ for some integer $k$.
For $k = 1$, $m = 8$ (which is a multiple of $8$).
For $k = 2$, $m = 14$ (which is NOT a multiple of $8$).
Thus, statement (C) is FALSE.

Evaluating Option (D):
For $z \in H$, let $z = x + iy$ where $y > 0$.
$\dfrac{az + b}{cz + d} = \dfrac{-1}{z + 1} = \dfrac{-1}{(x + 1) + iy}$
Multiplying the numerator and denominator by the conjugate $(x + 1) - iy$:
$\dfrac{-1}{(x + 1) + iy} \times \dfrac{(x + 1) - iy}{(x + 1) - iy} = \dfrac{-(x + 1) + iy}{(x + 1)^2 + y^2}$
The imaginary part of this new complex number is $\dfrac{y}{(x + 1)^2 + y^2}$.
Since $y > 0$ and $(x + 1)^2 + y^2 > 0$, the imaginary part is strictly positive.
Therefore, $\dfrac{az + b}{cz + d} \in H$, making statement (D) TRUE.

Answer: If $\omega = \dfrac{-1 + i\sqrt{3}}{2}$, then $\dfrac{a\omega + b}{c\omega + d} = \omega$; If $z \in H$, then $\dfrac{az + b}{cz + d} \in H$