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Question #41430

Difficulty: 8/10 | 🟢 Free

The Question

Let $z = e^{ix}$ . Evaluate the definite integral: $I = \int_{0}^{\pi} \text{Im} \left( \text{Log} \left( z + z^2 \right) \right) dx$ where $\text{Log}(w)$ denotes the principal branch of the complex logarithm, such that $\text{Im}(\text{Log}(w)) \in (-\pi, \pi]$ .

Options

A.
$\frac{\pi^2}{12}$
B.
$-\frac{\pi^2}{4}$
C.
$\frac{5\pi^2}{12}$
D.
$\frac{\pi^2}{6}$

Check Your Answer

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

Definite Integral Properties (King's Property)|De Moivre’s Theorem Applications|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

This question tests the application of De Moivre's theorem and the strict domain constraints of the principal complex logarithm.

A common trap is blindly applying King's property: $I = \int_0^\pi \text{Arg}(e^{i(\pi-x)} + e^{i2(\pi-x)}) dx = \int_0^\pi \text{Arg}(-e^{-ix} + e^{-i2x}) dx$ .

Students then incorrectly assume $\text{Arg}(z_1) + \text{Arg}(z_2) = \text{Arg}(z_1 z_2)$ to add the integrals, yielding $2I = \int_0^\pi \text{Arg}(-2i\sin x) dx = \int_0^\pi (-\pi/2) dx = -\pi^2/2 \implies I = -\pi^2/4$ .

This fails because the argument product rule requires modulo $2\pi$ adjustments, which King's property masks if applied algebraically without branch cut analysis.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Definite Integral Properties (King's Property)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free