JEE Maths›Complex Numbers & Quadratic Equations›Visual Solution
Visual SolutionPYQ 2024Standard
Question
If ∣z−2∣=∣z+2∣|z - 2| = |z + 2|, then zz lies on:
(A)The real axis
(B)The imaginary axis
(C)A circle of radius 2
(D)A circle of radius 4
Solution Path
∣z−2∣=∣z+2∣|z-2| = |z+2| means equidistant from 22 and −2-2. Perpendicular bisector of the segment is Re(z)=0\text{Re}(z) = 0, i.e., the imaginary axis.
01Question Setup
1/4
Locus of zz equidistant from 22 and −2-2 on the Argand plane.
∣z−2∣=∣z+2∣|z - 2| = |z + 2|
02Geometric Meaning
2/4
Plot 22 and −2-2 on the real axis. ∣z−a∣|z - a| represents the distance from zz to the point aa.
dist(z,2)=dist(z,−2)\text{dist}(z, 2) = \text{dist}(z, -2)
03Perpendicular BisectorKEY INSIGHT
3/4
The set of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.
Re(z)=0\text{Re}(z) = 0
04Final Answer
4/4
The perpendicular bisector of the segment from −2-2 to 22 is the imaginary axis.
Imaginary axis\boxed{\text{Imaginary axis}} - Answer (B)
Concepts from this question2 concepts unlocked
★

Modulus Squared via Components

EASY

|a + bi|^2 = a^2 + b^2 - no square root needed

∣z∣2=z⋅zˉ=x2+y2|z|^2 = z \cdot \bar{z} = x^2 + y^2

Final step in many JEE problems. Students waste time square-rooting when they only need |z|^2

Distance problemsSum/product modulus\alpha^2 + \beta^2 type problems
Practice (11 Qs) →

Conjugate Symmetry on Argand Plane

STANDARD

Conjugate pairs are reflections across the real axis

z=x+iy  ⟹  zˉ=x−iyz = x + iy \implies \bar{z} = x - iy

Recognizing conjugate pairs halves the computation - products simplify dramatically

Locus problemsGeometric interpretationSymmetric root problems
Practice (11 Qs) →
Want more practice?
Try more PYQs from this chapter →