Class JEE Mathematics Practice Q #712
KNOWLEDGE BASED
APPLY
4 Marks 2023 MCQ SINGLE
If the complex number z satisfies |z| = 1 and z ≠ -1, then z/(1+z^2) is purely:
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Correct Answer: Real
Explanation
Let z = e^(iθ). Then z/(1+z^2) = e^(iθ) / (1 + e^(2iθ)). Simplifying this using Euler formula results in a real number.

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Step-by-Step Solution

Let z = cos(θ) + i sin(θ), since |z| = 1.

Then z2 = cos(2θ) + i sin(2θ).

So, 1 + z2 = 1 + cos(2θ) + i sin(2θ) = 2cos2(θ) + i(2sin(θ)cos(θ)) = 2cos(θ)[cos(θ) + i sin(θ)] = 2cos(θ)z.

Therefore, z / (1 + z2) = z / (2cos(θ)z) = 1 / (2cos(θ)) = (1/2)sec(θ), which is a real number.

Correct Answer: Real<\/strong>

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the properties of complex numbers and their modulus to determine the nature of the given expression.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concepts of complex numbers, modulus, and purely real or imaginary numbers. It's not just recalling facts but applying the definitions and properties.
Syllabus Audit: In the context of JEE, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of complex number properties, a core topic in the JEE syllabus.

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