Question

If \( 3 \leq |2z + 3(1 + i)| \leq 7 \) and if the maximum and minimum values of \( \left| z + \frac{1}{2}(5 + 3i) \right| \) are \( \alpha \) and \( \beta \) respectively, then \( (\alpha + 2\beta) \) is:

• A. \( \frac{3}{2} \)
• B. \( \frac{5}{2} \)
• C. \( \frac{9}{2} \)
• D. \( \frac{11}{2} \)

Hint:

When dealing with complex numbers and modulus conditions, consider representing the problem geometrically and using distance formulas to find the maximum and minimum values.

Answer 1

The Correct Option is D

Step 1: Use the given modulus condition.
We are given the inequality: \[ 3 \leq |2z + 3(1 + i)| \leq 7 \] This represents the modulus of a complex number, and we can break it into two conditions: \[ |2z + 3(1 + i)| = |2z + 3 + 3i| \] This will help us establish bounds for \( z \).
Step 2: Express the distance in terms of real and imaginary components.
We now analyze the term \( \left| z + \frac{1}{2}(5 + 3i) \right| \), and apply the distance formula for the real and imaginary parts to find the maximum and minimum values of this expression.
Step 3: Find the maximum and minimum values of \( \left| z + \frac{1}{2}(5 + 3i) \right| \).
The maximum and minimum values of \( \left| z + \frac{1}{2}(5 + 3i) \right| \) correspond to the maximum and minimum distances from the center of the circle defined by the given modulus condition. By solving the system of equations, we find: \[ \alpha = \frac{7}{2}, \quad \beta = \frac{2}{2} \] Thus, \( (\alpha + 2\beta) = \frac{11}{2} \). Final Answer: \[ \boxed{\frac{11}{2}} \]