Complex Numbers
Imaginary Part Zero Condition — Circle Equation
nta_pyq_2023_apr
Grade 11
Question:
Let $S=\left\{z\in\mathbb{C}\setminus\{i,2i\}:\ \dfrac{z^2+8iz-15}{z^2-3iz-2}\in\mathbb{R}\right\}$. If $\alpha-\dfrac{13i}{11}\in S,\ \alpha\in\mathbb{R}\setminus\{0\}$, then $242\alpha^2$ is equal to
Step-by-Step Solution
Key Concept: Set $z=x+iy$ and compute Im$\left(\frac{z^2+8iz-15}{z^2-3iz-2}\right)=0$. After simplification, this reduces to $11x^2+11y^2+26y-61=0$ (a circle equation).
Imaginary part $=0$ gives $11x^2+11y^2+26y-61=0$. With $y=-\frac{13}{11}$: $121\alpha^2=840$. $242\alpha^2=1680$.
Correct Answer: 1680