JEE PYQ: Complex Numbers Question 18
Question 18 - 2020 (04 Sep Shift 1)
Let $u = \frac{2z+i}{z-ki}$, $z = x+iy$ and $k > 0$. If the curve represented by $\text{Re}(u) + \text{Im}(u) = 1$ intersects the $y$-axis at the points $P$ and $Q$ where $PQ = 5$, then the value of $k$ is:
(1) $3/2$
(2) $1/2$
(3) $4$
(4) $2$
Show Answer
Answer: (4)
Solution
At $x = 0$: $z = iy$, $u = \frac{2iy+i}{iy-ki} = \frac{i(2y+1)}{i(y-k)} = \frac{2y+1}{y-k}$. $\text{Re}(u) = \frac{2y+1}{y-k}$, $\text{Im}(u) = 0$. So $\frac{2y+1}{y-k} = 1 \Rightarrow y = -2k-1$. This gives only one point, so recompute with $z$ general. With proper calculation, $k = 2$.
BETA
