The value of k for which the inequality `| Re (z) | + | Im (z)| leq lambda |z|` is true for all `z in C` , is

Options

A. 2
B. `sqrt(2)`
C. 1
D. none of these

Explanation

To solve the inequality \( | \text{Re}(z) | + | \text{Im}(z) | \leq \lambda |z| \) for all \( z \in \mathbb{C} \), we can follow these steps:

### Step-by-Step Solution:

1. **Express \( z \) in polar form**:
Let \( z = r(\cos \theta + i \sin \theta) \), where \( r = |z| \) and \( \theta \) is the argument of \( z \).

2. **Calculate \( | \text{Re}(z) | \) and \( | \text{Im}(z) | \)**:
- The real part of \( z \) is \( \text{Re}(z) = r \cos \theta \).
- The imaginary part of \( z \) is \( \text{Im}(z) = r \sin \theta \).
- Thus, we have:
\[ | \text{Re}(z) | + | \text{Im}(z) | = |r \cos \theta| + |r \sin \theta| = r |\cos \theta| + r |\sin \theta| = r (|\cos \theta| + |\sin \theta|). \]

3. **Substitute into the inequality**:
The inequality becomes:
\[ r (|\cos \theta| + |\sin \theta|) \leq \lambda r. \]
Since \( r \) is positive (as it is the modulus), we can divide both sides by \( r \):
\[ |\cos \theta| + |\sin \theta| \leq \lambda. \]

4. **Find the maximum value of \( |\cos \theta| + |\sin \theta| \)**:
To find the maximum value of \( |\cos \theta| + |\sin \theta| \), we can use the Cauchy-Schwarz inequality:
\[ |\cos \theta| + |\sin \theta| \leq \sqrt{(\cos^2 \theta + \sin^2 \theta)(1^2 + 1^2)} = \sqrt{1 \cdot 2} = \sqrt{2}. \]
This maximum occurs when \( |\cos \theta| = |\sin \theta| \), which happens at \( \theta = \frac{\pi}{4} + n\frac{\pi}{2} \) for \( n \in \mathbb{Z} \).

5. **Set \( \lambda \) equal to the maximum value**:
For the inequality to hold for all \( z \in \mathbb{C} \), we must have:
\[ \lambda \geq \sqrt{2}. \]
Therefore, the smallest value of \( \lambda \) that satisfies this condition is:
\[ \lambda = \sqrt{2}. \]

### Final Answer:
The value of \( \lambda \) for which the inequality holds for all \( z \in \mathbb{C} \) is \( \sqrt{2} \).