Question

In Argand plane, no value of \( \sqrt[3]{1-i\sqrt{3}} \) lie in

• A. First quadrant
• B. Second quadrant
• C. Third quadrant
• D. Fourth quadrant

Hint:

The nth roots of a complex number form a regular polygon centered at the origin. For cube roots, they are \( 120^\circ \) apart. Once you find one argument, simply add \( 120^\circ \) to find the others.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We need to find the three cube roots of the given complex number and determine their positions (quadrants) in the Argand plane.
Step 2: Detailed Explanation:

Let \( z = 1 - i\sqrt{3} \). 1. Find Polar Form: Modulus \( r = \sqrt{1^2 + (-\sqrt{3})^2} = 2 \). Argument \( \theta \): Since real part \(>0 \) and imaginary part \(<0 \), it is in the 4th quadrant. \( \theta = \tan^{-1}(-\sqrt{3}/1) = -\frac{\pi}{3} \). So, \( z = 2 e^{-i\pi/3} \). 2. Find Cube Roots: The roots are given by \( w_k = r^{1/3} e^{i(\frac{\theta + 2k\pi}{3})} \) for \( k = 0, 1, 2 \). Radius of roots \( R = 2^{1/3} \). The quadrant depends only on the argument. Arguments \( \phi_k = \frac{-\pi/3 + 2k\pi}{3} = -\frac{\pi}{9} + \frac{2k\pi}{3} \). * For \( k = 0 \): \( \phi_0 = -\frac{\pi}{9} = -20^\circ \). This lies in the Fourth Quadrant (\( -90^\circ<\phi<0^\circ \)). * For \( k = 1 \): \( \phi_1 = -\frac{\pi}{9} + \frac{6\pi}{9} = \frac{5\pi}{9} = 100^\circ \). This lies in the Second Quadrant (\( 90^\circ<\phi<180^\circ \)). * For \( k = 2 \): \( \phi_2 = -\frac{\pi}{9} + \frac{12\pi}{9} = \frac{11\pi}{9} = 220^\circ \). This lies in the Third Quadrant (\( 180^\circ<\phi<270^\circ \)). 3. Conclusion: The roots lie in Quadrants II, III, and IV. No root lies in the First Quadrant.
Step 4: Final Answer:

No value lies in the First quadrant.