Question

If $|z + 1| < |z - 1|$, then $z$ lies:

• A. on the x-axis
• B. on the y-axis
• C. in the region $x < 0$
• D. in the region $y > 0$
• E. in the region $x > y$

Hint:

You can verify algebraically: \[ (x+1)^2 + y^2 < (x-1)^2 + y^2 \Rightarrow 4x < 0 \Rightarrow x < 0 \] Always remember: such inequalities divide the plane into two half-planes.

Answer 1

The Correct Option is C

Concept: The modulus $|z-a|$ represents the distance between point $z$ and point $a$ in the complex plane. The inequality $|z-a| < |z-b|$ defines the region of points closer to $a$ than to $b$.

Step 1: Interpret geometrically.
Let $z = x + iy$. \[ |z + 1| = |z - (-1)| \Rightarrow \text{distance from } (-1,0) \] \[ |z - 1| \Rightarrow \text{distance from } (1,0) \] Thus, the inequality means points closer to $(-1,0)$ than $(1,0)$.

Step 2: Find the boundary line.
Points equidistant from $(-1,0)$ and $(1,0)$ lie on the perpendicular bisector: \[ x = 0 \quad (\text{y-axis}) \]

Step 3: Determine the region.
Since $(-1,0)$ lies on the left side of the y-axis, the required region is: \[ x < 0 \]