Question

\( \omega \) is a complex cube root of unity and \( Z \) is a complex number satisfying \( |Z-1| \le 2 \). The possible values of \( r \) such that \( |Z-1| \le 2 \) and \( |\omega Z - 1 - \omega^2| = r \) have no common solution are

• A. \( 0 \le r \le 4 \)
• B. \( r = |\omega| \) only
• C. \( r>4 \)
• D. \( 1<r<2 \)

Hint:

Always simplify complex equations like \( |\omega Z + \dots| \) by factoring out \( |\omega| \) to reveal the standard circle form \( |Z - z_0| = r \). Drawing a diagram with centers and radii often makes the intersection conditions obvious.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

The problem involves the geometric interpretation of complex numbers. The inequality \( |Z-z_0| \le R \) represents a disk centered at \( z_0 \) with radius \( R \). The equation \( |Z-z_1| = r \) represents a circle centered at \( z_1 \) with radius \( r \). We need to find \( r \) such that the disk and the circle do not intersect.
Step 2: Detailed Explanation:

1. Analyze the first region: \( |Z-1| \le 2 \) represents a solid disk centered at \( C_1(1, 0) \) with radius \( R_1 = 2 \). Let's denote the set of points in this disk as \( S_1 \). 2. Analyze the second equation: Given: \( |\omega Z - 1 - \omega^2| = r \). Factor out \( \omega \) (knowing \( |\omega| = 1 \)): \[ |\omega(Z - \frac{1}{\omega} - \omega)| = r \] Since \( \frac{1}{\omega} = \bar{\omega} = \omega^2 \) and \( \omega + \omega^2 = -1 \): \[ |Z - (\omega^2 + \omega)| = r \] \[ |Z - (-1)| = r \implies |Z + 1| = r \] This represents a circle centered at \( C_2(-1, 0) \) with radius \( r \). Let's denote this set of points as \( S_2 \). 3. Condition for "No Common Solution": We need the intersection of the disk \( S_1 \) and the circle \( S_2 \) to be empty. The center of the circle \( C_2(-1, 0) \) lies on the boundary of the disk \( S_1 \), because the distance between \( C_1(1,0) \) and \( C_2(-1,0) \) is: \[ d = |1 - (-1)| = 2 \] Since the distance \( d \) equals the radius \( R_1 \) of the disk, the center \( C_2 \) is exactly on the edge of the disk. 4. Finding the range of distances from \( C_2 \) to any point in \( S_1 \): Let \( Z \) be any point in the disk \( |Z-1| \le 2 \). We want to find the range of values for \( |Z+1| \). * Minimum value: Since \( -1 \) is on the boundary of the disk, the closest point in the disk to \( -1 \) is \( -1 \) itself. So, \( |Z+1|_{\min} = 0 \). * Maximum value: The point in the disk farthest from \( -1 \) lies on the diameter passing through \( -1 \) and \( 1 \). The disk extends from \( 1-2 = -1 \) to \( 1+2 = 3 \) on the real axis. The distance from \( -1 \) to \( 3 \) is \( |3 - (-1)| = 4 \). So, for any \( Z \) in the disk, \( |Z+1| \in [0, 4] \). 5. Conclusion: The circle \( |Z+1| = r \) intersects the disk if \( r \) falls within the range of distances \( [0, 4] \). For there to be no common solution, \( r \) must lie outside this interval. Since \( r \) is a radius (must be positive), we must have: \[ r>4 \]
Step 4: Final Answer:

The possible values are \( r>4 \).