Complex Numbers & Matrices — Möbius Transformation
PYP_JEE_ADV_2026_P2
Grade None
Question:
Let $\mathbb{R}$ denote the set of all real numbers and let $i=\sqrt{-1}$. Consider the matrices
$$S=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \quad\text{and}\quad T=\begin{bmatrix}1&1\\0&1\end{bmatrix}.$$
Let $a,b,c,d$ be real numbers such that
$$ST=\begin{bmatrix}a&b\\c&d\end{bmatrix}.$$
Let
$$H=\{x+iy:\ x,y\in\mathbb{R}\text{ and }y>0\}.$$
Then which of the following statements is (are) TRUE?
$\dfrac{b+ia}{d+ic}=i$
If $\omega=\dfrac{-1+i\sqrt{3}}{2}$, then $\dfrac{a\omega+b}{c\omega+d}=\omega$
If $m$ is an integer greater than $2$ such that $(ST)^2=(ST)^m$, then $m$ is an integer multiple of $8$
If $z\in H$, then $\dfrac{az+b}{cz+d}\in H$
Step-by-Step Solution
Key Concept: The matrix $ST$ induces a Möbius transformation $z\mapsto\frac{-1}{z+1}$. The upper half-plane $H$ is preserved under Möbius transformations with real matrix entries and positive determinant.
**Step 1: Compute ST**
$ST=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}0&-1\\1&1\end{bmatrix}$. So $a=0,b=-1,c=1,d=1$.
**Step 2: Check (A)**
$\frac{b+ia}{d+ic}=\frac{-1+0}{1+i}=\frac{-1}{1+i}=\frac{-1+i}{2}\neq i$. FALSE ✗
**Step 3: Check (B)**
$\frac{a\omega+b}{c\omega+d}=\frac{-1}{\omega+1}$. $\omega+1=\frac{1+i\sqrt{3}}{2}$. $\frac{-1}{\frac{1+i\sqrt{3}}{2}}=\frac{-2}{1+i\sqrt{3}}=\frac{-2(1-i\sqrt{3})}{4}=\frac{-1+i\sqrt{3}}{2}=\omega$. TRUE ✓
**Step 4: Check (C): order of ST**
$(ST)^3=\begin{bmatrix}0&-1\\1&1\end{bmatrix}^2\cdot ST=\begin{bmatrix}-1&-1\\1&0\end{bmatrix}\begin{bmatrix}0&-1\\1&1\end{bmatrix}=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}=-I$. So $(ST)^6=I$, order $=6$. $(ST)^2=(ST)^m$ iff $m\equiv2\pmod6$. So $m\in\{8,14,20,...\}$. $m=14$ is not a multiple of 8. FALSE ✗
**Step 5: Check (D): H is preserved**
$\frac{az+b}{cz+d}=\frac{-1}{z+1}$. $\text{Im}\left(\frac{-1}{z+1}\right)=\frac{\text{Im}(z)}{|z+1|^2}>0$ since $\text{Im}(z)>0$. So the image lies in $H$. TRUE ✓
Correct Answer: B, D