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Conjugate And Reciprocal

Hardmathematics

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?

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About This Question

Subject
mathematics
Chapter
complex numbers and quadratic equations
Topic
conjugate and reciprocal
Difficulty
Hard
Year
2025
Tags
advanced-calculus-drillconjugatereality-conditionreciprocallocus

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.

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