UPSC NDA I 2016 Mathematics Solved Paper

Concept:Use the geometric interpretation of modulus to find the complex number $z$ from given conditions, then evaluate the required expression.Explanation:From $\left|\frac{z-4}{z-8}\right| = 1$, we have $|z-4| = |z-8|$.This means $z$ lies on the perpendicular bisector of the points $4$ and $8$ on the real axis.The midpoint is $6$, so $z = 6 + iy$ where $y$ is real.Now use the second condition: $\left|\frac{z}{z-2}\right| = \frac{3}{2}$.Substitute $z = 6 + iy$: $|z|^2 = 36 + y^2$, $|z-2|^2 = |4+iy|^2 = 16 + y^2$.Thus $\frac{36+y^2}{16+y^2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.Cross‑multiply: $4(36+y^2) = 9(16+y^2)$ gives $144+4y^2 = 144+9y^2$.This simplifies to $4y^2 = 9y^2$, so $y^2 = 0$, hence $y=0$.Therefore $z = 6$ (purely real).Now compute the required value: $\left|\frac{z-6}{z+6}\right| = \left|\frac{6-6}{6+6}\right| = \left|\frac{0}{12}\right| = 0$.Answer:$0$