Question

Given that \( i = \sqrt{-1} \). The value of \[ \lim_{z \to e^{i \pi/3}} \frac{z^3 + 1}{z^4 + z^2 + 1} \] is

• A. \( \frac{3}{4} + i \frac{\sqrt{3}}{4} \)
• B. \( \frac{3}{4} - i \frac{\sqrt{3}}{4} \)
• C. \( -\frac{3}{4} + i \frac{\sqrt{3}}{4} \)
• D. \( -\frac{3}{4} - i \frac{\sqrt{3}}{4} \)

Hint:

When dealing with limits resulting in indeterminate forms like \( 0/0 \), apply L'Hopital's Rule by differentiating the numerator and denominator separately.

Answer 1

The Correct Option is B

We are given the limit expression: \[ \lim_{z \to e^{i \pi/3}} \frac{z^3 + 1}{z^4 + z^2 + 1} \] To find the value of this limit, we first substitute \( z = e^{i \pi/3} \) into the expression. Start by evaluating the components in the numerator and denominator. 1. Numerator: \[ z^3 + 1 = \left( e^{i \pi/3} \right)^3 + 1 = e^{i \pi} + 1 = -1 + 1 = 0 \] 2. Denominator: \[ z^4 + z^2 + 1 = \left( e^{i \pi/3} \right)^4 + \left( e^{i \pi/3} \right)^2 + 1 = e^{i 4\pi/3} + e^{i 2\pi/3} + 1 \] Using Euler's formula for the complex exponentials: \[ e^{i 4\pi/3} = -\frac{1}{2} - i \frac{\sqrt{3}}{2}, e^{i 2\pi/3} = -\frac{1}{2} + i \frac{\sqrt{3}}{2} \] Therefore: \[ e^{i 4\pi/3} + e^{i 2\pi/3} + 1 = \left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) + \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) + 1 = -1 + 1 = 0 \] Since both the numerator and denominator tend to zero, we apply L'Hopital's Rule, which involves differentiating the numerator and denominator separately with respect to \( z \). - Numerator derivative: \[ \frac{d}{dz}(z^3 + 1) = 3z^2 \] At \( z = e^{i \pi/3} \), we get: \[ 3\left( e^{i \pi/3} \right)^2 = 3 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -\frac{3}{2} + i \frac{3\sqrt{3}}{2} \] - Denominator derivative: \[ \frac{d}{dz}(z^4 + z^2 + 1) = 4z^3 + 2z \] At \( z = e^{i \pi/3} \), we get: \[ 4 \left( e^{i \pi/3} \right)^3 + 2 \left( e^{i \pi/3} \right) = 4 \left( -1 \right) + 2 \left( e^{i \pi/3} \right) = -4 + 2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \] Simplifying: \[ -4 - 1 + i \sqrt{3} = -5 + i \sqrt{3} \] Now, the limit is: \[ \lim_{z \to e^{i \pi/3}} \frac{-\frac{3}{2} + i \frac{3\sqrt{3}}{2}}{-5 + i \sqrt{3}} \] By simplifying, the value of the limit becomes: \[ \frac{3}{4} - i \frac{\sqrt{3}}{4} \] Thus, the correct answer is (B).
Final Answer: (B) \( \frac{3}{4} - i \frac{\sqrt{3}}{4} \)