Before You Begin

Question #40463

Difficulty: 8/10 | 🟢 Free

The Question

Let $z$ be a non-zero complex number and $w = u + iv$ ( $u, v \in \mathbb{R}$ ) be its principal square root, defined such that the argument $\theta$ of $z$ satisfies $\theta \in (-\pi, \pi]$ . A vector $\vec{d} = u\hat{i} + v\hat{j} + \sqrt{2}\hat{k}$ has direction cosines $l, m, n$ along the positive x, y, and z axes respectively. If $n = \frac{1}{2}$ and $m = -\frac{\sqrt{3}}{4}$ , then $z$ is equal to:

Options

A.
$3 - 3\sqrt{3}i$
B.
$3 + 3\sqrt{3}i$
C.
$-3 - 3\sqrt{3}i$
D.
$-3 + 3\sqrt{3}i$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Square Root of Complex Numbers|Direction Cosines constraint|The Quadrant Adjustment Logic

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of 3D geometry (direction cosines) and complex numbers (principal square root).

The specific trap lies in the extraction of the real part $u$ from the direction cosine $l$ .

Since $l^2 = 9/16$ , students often arbitrarily choose a sign or assume both are valid.

However, the definition of the principal square root of a complex number $z$ strictly restricts its argument to $(-\pi/2, \pi/2]$ , which mathematically mandates that its real part $u$ must be non-negative.

Failing to apply this quadrant adjustment logic leads to choosing $u < 0$ , which incorrectly yields the conjugate $3 + 3\sqrt{3}i$ as the answer.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Complex Numbers and Quadratic Equations - Class 11
Concept	Square Root of Complex Numbers
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free