Question

If \( \omega \) is a complex cube root of unity and if \( Z \) is a complex number satisfying \( |Z - 1| \leq 2 \) and \[ |\omega^2 Z - 1 - \omega| = a, \] then the set of possible values of \( a \) is:

• A. \( 0 \leq a \leq 2 \)
• B. \( \frac{1}{2} \leq a \leq \frac{\sqrt{3}}{2} \)
• C. \( |\omega| \leq a \leq \frac{\sqrt{3}}{2} + 2 \)
• D. \( 0 \leq a \leq 4 \)

Hint:

The Triangle Inequality states that for any complex numbers \(z_1\) and \(z_2\), \(|z_1 + z_2| \le |z_1| + |z_2|\) and \(|z_1 + z_2| \ge |z_1| - |z_2|\).

Answer 1

The Correct Option is D

The condition \( |Z - 1| \leq 2 \) means \( Z \) lies in a disk centered at 1 with radius 2. 

Since \( \omega^3 = 1 \) and \( \omega \neq 1 \), \( \omega^3 - 1 = 0 \), which factors as:

\[ (\omega - 1)(\omega^2 + \omega + 1) = 0 \]

Since \( \omega \neq 1 \), \( \omega^2 + \omega + 1 = 0 \), so \( \omega^2 = -1 - \omega \).

Then,

\[ |\omega^2 Z - 1 - \omega| = |(-1 - \omega) Z - 1 - \omega| = |-(1 + \omega) Z - (1 + \omega)| = |1 + \omega| \cdot |Z + 1| = |Z + 1|. \]

By the Triangle Inequality,

\[ |Z + 1| = |Z - 1 + 2| \leq |Z - 1| + 2 \leq 2 + 2 = 4. \]

Equality occurs when \( Z = 3 \).

Also, by the Triangle Inequality,

\[ |Z + 1| = |Z - 1 + 2| \geq |2| - |Z - 1| \geq 2 - 2 = 0. \]

Equality occurs when \( Z = -1 \).

Hence, the set of possible values of \( a \) is:

\[ \boldsymbol{0 \leq a \leq 4.} \]