Complex Numbers
Complex Numbers
star_batch_jee_advanced_2025
Grade 11
Question:
Locus of $z$, if $\arg(z - (1+i)) = \begin{cases} \frac{3\pi}{4}, & \text{when } |z| \leq |z-2| \\ -\frac{\pi}{4}, & \text{when } |z| > |z-2| \end{cases}$ is:
a pair of straight lines passing through (2, 0)
a pair of straight lines passing through (2, 0), (1, 1)
a line segment
a set of two rays
Step-by-Step Solution
Key Concept: The locus consists of two rays (half-lines) emanating from $(1,1)$ in opposite directions along the line $x+y=2$, with the boundary condition at $x=1$ separating the two rays.
We analyze this problem by first understanding the conditions. The locus is divided by the perpendicular bisector of the segment from origin to $(2,0)$, which is the line $x=1$. For $|z| \leq |z-2|$ (points on or left of $x=1$): $\arg(z-(1+i)) = \frac{3\pi}{4}$ gives a ray from $(1,1)$ in direction $e^{i3\pi/4}$, which is the line $y-1 = -(x-1)$ or $x+y=2$ for $x \leq 1$. For $|z| > |z-2|$ (points right of $x=1$): $\arg(z-(1+i)) = -\frac{\pi}{4}$ gives a ray from $(1,1)$ in direction $e^{-i\pi/4}$, which is the line $y-1 = -(x-1)$ or $x+y=2$ for $x \geq 1$. Both conditions actually define rays emanating from the point $(1,1)$ along the line $x+y=2$, with the first ray going left and the second going right.
Correct Answer: 4