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Question #40494

Difficulty: 8/10 | 🟢 Free

The Question

Let z1z_1 and z2z_2 be the roots of the complex quadratic equation z2−(1+3i)z−2+2i=0z^2 - (1+3i)z - 2 + 2i = 0. The value of ∣z1∣+∣z2∣|z_1| + |z_2| is:

Options

  • A.

    2+22 + \sqrt{2}

  • B.

    2^{7/4}

  • C.

    10\sqrt{10}

  • D.

    6

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Conjugate Root Theorem|Quadratic Equations with Complex Coefficients|Modulus Function Symmetry

Why This Question Is Tricky

Common Mistakes

This question tests the boundary conditions of the Conjugate Root Theorem.

A common reflex among students is to assume that the roots of any quadratic equation are complex conjugates, which implies ∣z1∣=∣z2∣|z_1| = |z_2|.

If applied here, students would use the product of roots ∣z1z2∣=∣−2+2i∣=22|z_1 z_2| = |-2+2i| = 2\sqrt{2}, deduce ∣z1∣=23/4|z_1| = 2^{3/4}, and incorrectly conclude the sum is 27/42^{7/4}.

The trap penalizes rote memorization by using complex coefficients, breaking the symmetry and forcing a rigorous application of the quadratic formula and modulus definition.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterComplex Numbers and Quadratic Equations - Class 11
ConceptConjugate Root Theorem
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree