If `z=(3+7i)(lambda+imu)," when " lambda,mu in I-{0} " and " i=sqrt(-1)`, is purely imaginary then minimum value of `abs(z)^(2)` is

Options

A. 0
B. 58
C. `3364/3`
D. 3364

Answer

To solve the problem, we need to find the minimum value of \( |z|^2 \) where \( z = (3 + 7i)(\lambda + i\mu) \) and \( \lambda, \mu \in \mathbb{I} \setminus \{0\} \), and \( z \) is purely imaginary.

### Step-by-Step Solution:

1. **Expand the expression for \( z \)**:
\[ z = (3 + 7i)(\lambda + i\mu) = 3\lambda + 3i\mu + 7i\lambda + 7i^2\mu \]
Since \( i^2 = -1 \), we can rewrite this as:
\[ z = 3\lambda - 7\mu + i(3\mu + 7\lambda) \]

2. **Identify the real and imaginary parts**:
The real part of \( z \) is \( 3\lambda - 7\mu \) and the imaginary part is \( 3\mu + 7\lambda \).

3. **Set the real part to zero for \( z \) to be purely imaginary**:
\[ 3\lambda - 7\mu = 0 \]
This implies:
\[ 3\lambda = 7\mu \quad \Rightarrow \quad \lambda = \frac{7}{3}\mu \]

4. **Substitute \( \lambda \) in terms of \( \mu \)**:
Substitute \( \lambda = \frac{7}{3}\mu \) into the imaginary part:
\[ z = 3\mu + 7\left(\frac{7}{3}\mu\right) = 3\mu + \frac{49}{3}\mu = \frac{3\mu + 49\mu}{3} = \frac{52\mu}{3}i \]

5. **Calculate the modulus of \( z \)**:
The modulus of \( z \) is:
\[ |z| = \left|\frac{52\mu}{3}i\right| = \frac{52}{3}|\mu| \]

6. **Find \( |z|^2 \)**:
\[ |z|^2 = \left(\frac{52}{3}|\mu|\right)^2 = \frac{2704}{9}|\mu|^2 \]

7. **Minimize \( |z|^2 \)**:
Since \( \mu \) is a non-zero integer, the minimum value of \( |\mu|^2 \) is \( 1 \) (when \( \mu = 1 \) or \( \mu = -1 \)):
\[ |z|^2_{\text{min}} = \frac{2704}{9} \cdot 1 = \frac{2704}{9} \]

### Final Answer:
The minimum value of \( |z|^2 \) is \( \frac{2704}{9} \).