JEE Advanced 2018 — Paper 2 — Question 32

Write the equation as $s z + t \bar{z} = -r$.

Taking conjugate gives $\bar{s} \bar{z} + \bar{t} z = -\bar{r}$.

This is a linear system in $z$ and $\bar{z}$. The determinant of the system is $\Delta = |s|^2 - |t|^2$. If $\Delta \neq 0$, the system has a unique solution for $z$, so L has exactly one element. If $\Delta = 0$ (i.e., $|s| = |t|$), the equations are dependent; L is either empty or a line (infinite points).

Example: $s = t = 1, r = i$ gives no solution. Hence B is false. For statement A: If L has exactly one element, then $\Delta \neq 0$ so $|s| \neq |t|$. Thus A is true. Statement C: The given equation reduces to a linear equation in $x, y$, so L is empty, a point, or a line.

The set $\{z : |z - 1 + i| = 5\}$ is a circle.

A line and a circle intersect at most 2 points; a point and a circle at most 1.

Thus $|L \cap \text{circle}| \leq 2$. So C is true. Statement D: If L has more than one element, it cannot be a single point, so it must be a line, which has infinitely many elements.

Hence D is true. Therefore, the correct statements are A, C, and D.