Conjugate And Reciprocal
If z is a non-zero complex number with |z| = 2 and the property that z plus its reciprocal 1/z is real, then z must lie on which set in the Argand plane?
Select the correct option:
Solution
The real axis only
A complex number w is real exactly when it equals its own conjugate, a criterion repeatedly invoked in JEE Advanced. Let w = z + 1/z; then w real means z + 1/z = conjugate of (z + 1/z) = z-bar + 1/z-bar. Rearranging, z - z-bar = 1/z-bar - 1/z = (z - z-bar)/(z z-bar) = (z - z-bar)/|z|^2. So (z - z-bar)(1 - 1/|z|^2) = 0, meaning either z - z-bar = 0 (z real) or |z|^2 = 1 (z on the unit circle). Now apply the given constraint |z| = 2: since |z|^2 = 4 ≠ 1, the unit-circle branch is impossible, so the factor 1 - 1/|z|^2 = 3/4 is non-zero and we must have z - z-bar = 0. Hence z is real, and with |z| = 2 the only such points are z = 2 and z = -2, both on the real axis. Option a circle of radius 1 is excluded precisely because |z| = 2. Option circle of radius 2 confuses the given magnitude constraint with the reality locus; the reality condition is not met everywhere on |z| = 2. Option imaginary axis fails since a purely imaginary z = 2i gives z + 1/z = 2i - i/2 = (3/2)i, which is not real. Hence z lies on the real axis only. Plausibility check: z = 2 gives z + 1/z = 2 + 1/2 = 5/2, real, while z = 2i gives a non-real value, confirming that under |z| = 2 only the real-axis points qualify.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- mathematics
- Chapter
- complex numbers and quadratic equations
- Topic
- conjugate and reciprocal
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
The real axis only
A complex number w is real exactly when it equals its own conjugate, a criterion repeatedly invoked in JEE Advanced. Let w = z + 1/z; then w real means z + 1/z = conjugate of (z + 1/z) = z-bar + 1/z-bar. Rearranging, z - z-bar = 1/z-bar - 1/z = (z - z-bar)/(z z-bar) = (z - z-bar)/|z|^2. So (z - z-bar)(1 - 1/|z|^2) = 0, meaning either z - z-bar = 0 (z real) or |z|^2 = 1 (z on the unit circle). Now apply the given constraint |z| = 2: since |z|^2 = 4 ≠ 1, the unit-circle branch is impossible, so the factor 1 - 1/|z|^2 = 3/4 is non-zero and we must have z - z-bar = 0. Hence z is real, and with |z| = 2 the only such points are z = 2 and z = -2, both on the real axis. Option a circle of radius 1 is excluded precisely because |z| = 2. Option circle of radius 2 confuses the given magnitude constraint with the reality locus; the reality condition is not met everywhere on |z| = 2. Option imaginary axis fails since a purely imaginary z = 2i gives z + 1/z = 2i - i/2 = (3/2)i, which is not real. Hence z lies on the real axis only. Plausibility check: z = 2 gives z + 1/z = 2 + 1/2 = 5/2, real, while z = 2i gives a non-real value, confirming that under |z| = 2 only the real-axis points qualify.
This hard difficulty mathematics question is from the chapter complex numbers and quadratic equations, covering the topic of conjugate and reciprocal. It appeared in the 2025 exam.
Looking for more practice? Explore all mathematics questions or browse complex numbers and quadratic equations questions on RankGuru.