Modulus Of Complex Numbers
If z is a complex number satisfying the modulus relation |z - 3| = |z + 3|, then the point representing z must lie on which specific line in the Argand plane?
Select the correct option:
Solution
The imaginary axis
The modulus |z - a| represents the distance of the point z from the point a in the Argand plane, a geometric viewpoint central to JEE Advanced complex-number problems. The equation |z - 3| = |z + 3| states that z is equidistant from the points 3 and -3 on the real axis. The set of points equidistant from two fixed points is the perpendicular bisector of the segment joining them. Since 3 and -3 are symmetric about the origin on the real axis, their perpendicular bisector is the imaginary axis, the line x = 0. Writing z = x + iy confirms this: (x-3)^2 + y^2 = (x+3)^2 + y^2 gives -6x = 6x, so x = 0. Option real axis would require equidistance from points on the imaginary axis. Option y = x has no symmetry justification here. Option circle of radius 3 confuses distance equality with a fixed distance. Hence z lies on the imaginary axis. Plausibility check: any purely imaginary z = iy is clearly equidistant from 3 and -3 by symmetry, confirming the locus.
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About This Question
- Subject
- mathematics
- Chapter
- complex numbers and quadratic equations
- Topic
- modulus of complex numbers
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
The imaginary axis
The modulus |z - a| represents the distance of the point z from the point a in the Argand plane, a geometric viewpoint central to JEE Advanced complex-number problems. The equation |z - 3| = |z + 3| states that z is equidistant from the points 3 and -3 on the real axis. The set of points equidistant from two fixed points is the perpendicular bisector of the segment joining them. Since 3 and -3 are symmetric about the origin on the real axis, their perpendicular bisector is the imaginary axis, the line x = 0. Writing z = x + iy confirms this: (x-3)^2 + y^2 = (x+3)^2 + y^2 gives -6x = 6x, so x = 0. Option real axis would require equidistance from points on the imaginary axis. Option y = x has no symmetry justification here. Option circle of radius 3 confuses distance equality with a fixed distance. Hence z lies on the imaginary axis. Plausibility check: any purely imaginary z = iy is clearly equidistant from 3 and -3 by symmetry, confirming the locus.
This easy difficulty mathematics question is from the chapter complex numbers and quadratic equations, covering the topic of modulus of complex numbers. It appeared in the 2025 exam.
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