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Question #41748

Difficulty: 8/10 | 🟢 Free

The Question

Let the direction vector of a line in the Argand plane be represented by the complex number z0=−7+3iz_0 = -7 + 3i. The vector z0z_0 is orthogonally projected onto the boundary line of the region defined by the inequality ∣z−2∣≤∣z−2i∣|z - 2| \le |z - 2i|. If ww is the complex number representing this projection, find the principal value of the complex logarithm, Log(w)\text{Log}(w).

Options

  • A.

    32ln⁡2−i3π4\frac{3}{2}\ln 2 - i\frac{3\pi}{4}

  • B.

    32ln⁡2+iπ4\frac{3}{2}\ln 2 + i\frac{\pi}{4}

  • C.

    32ln⁡2+i5π4\frac{3}{2}\ln 2 + i\frac{5\pi}{4}

  • D.

    12ln⁡2−i3π4\frac{1}{2}\ln 2 - i\frac{3\pi}{4}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Orthogonal projection of a line|Geometry of Complex Inequalities|Argument and Principal Value|Inverse Function Periodicity

Why This Question Is Tricky

Common Mistakes

This question tests the orthogonal projection of vectors mapped to the complex plane and the rigorous definition of the principal complex logarithm.

The primary trap lies in the periodicity of the inverse tangent function; students often blindly compute tan⁡−1(−2/−2)=π/4\tan^{-1}(-2/-2) = \pi/4 (Option B) or use the non-principal argument 5π/45\pi/4 (Option C), ignoring the strict principal value range Arg(z)∈(−π,π]\text{Arg}(z) \in (-\pi, \pi].

Additionally, Option D punishes students who identify the correct projection line but fail to compute the exact projection length (modulus) of the vector.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptOrthogonal projection of a line
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree