Complex Numbers

Concept:Matrix multiplication and Möbius transformations preserving the upper half‑plane.Explanation:First compute $ST$: $$ST = \begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix} = \begin{bmatrix}0&-1\\1&1\end{bmatrix}.$$ Hence $a=0$, $b=-1$, $c=1$, $d=1$.Option A: $\dfrac{b+ia}{d+ic} = \dfrac{-1}{1+i} = \dfrac{-1+i}{2} \neq i$. So A is false.Option B: $\omega = \dfrac{-1+i\sqrt3}{2}$ satisfies $\omega^2+\omega+1=0$, so $\omega+1 = -\omega^2$. Then $$\frac{a\omega+b}{c\omega+d} = \frac{-1}{\omega+1} = \frac{-1}{-\omega^2} = \frac{1}{\omega^2} = \omega,$$ since $\omega^3=1$. So B is true.Option C: Compute $(ST)^2 = \begin{bmatrix}-1&-1\\1&0\end{bmatrix}$, $(ST)^3 = \begin{bmatrix}-1&0\\0&-1\end{bmatrix} = -I$. Hence $(ST)^6 = I$. If $(ST)^2 = (ST)^m$, then $m \equiv 2 \pmod{6}$, which is not necessarily a multiple of $8$. So C is false.Option D: For $z = x+iy$ with $y>0$, $$\frac{az+b}{cz+d} = \frac{-1}{z+1} = \frac{-(x+1)+iy}{(x+1)^2+y^2}.$$ Its imaginary part $\dfrac{y}{(x+1)^2+y^2}$ is positive. Thus the image stays in $H$. So D is true.Answer:Statements B and D are true.