If ` z = x + iy, , x , y ` real and ` | x| + | y| le lambda | z| " then " lambda ` is equal to

Options

A. `sqrt(3)`
B. `sqrt(2)`
C. 1
D. none

Answer

To solve the problem, we need to show that \( |x| + |y| \leq \lambda |z| \) where \( z = x + iy \) and \( |z| = \sqrt{x^2 + y^2} \). We will find the value of \( \lambda \).

### Step-by-Step Solution:

1. **Start with the definition of \( z \)**:
\[ z = x + iy \]
where \( x \) and \( y \) are real numbers.

2. **Calculate the modulus of \( z \)**:
\[ |z| = \sqrt{x^2 + y^2} \]

3. **Express \( |x| + |y| \)**:
We want to show that:
\[ |x| + |y| \leq \lambda |z| \]

4. **Using the Cauchy-Schwarz Inequality**:
According to the Cauchy-Schwarz inequality in the context of real numbers:
\[ (|x| + |y|)^2 \leq (1^2 + 1^2)(x^2 + y^2) = 2(x^2 + y^2) \]
Taking the square root of both sides gives:
\[ |x| + |y| \leq \sqrt{2} \sqrt{x^2 + y^2} \]

5. **Substituting the modulus of \( z \)**:
Since \( |z| = \sqrt{x^2 + y^2} \), we can substitute:
\[ |x| + |y| \leq \sqrt{2} |z| \]

6. **Identifying \( \lambda \)**:
From the inequality \( |x| + |y| \leq \sqrt{2} |z| \), we can deduce that:
\[ \lambda = \sqrt{2} \]

### Conclusion:
Thus, the value of \( \lambda \) is \( \sqrt{2} \).