If `z=x+iy,x,y` real , then `|x|+|y|lek|z|` where k is equal to :

Options

A. 1
B. `sqrt2`
C. `sqrt3`
D. None of these

Answer

To solve the problem, we need to show that if \( z = x + iy \) where \( x \) and \( y \) are real numbers, then \( |x| + |y| \leq k|z| \) for some constant \( k \). We will find the value of \( k \).

### Step-by-Step Solution:

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

2. **Use the triangle inequality**:
The triangle inequality states that for any complex numbers \( a \) and \( b \):
\[ |a + b| \leq |a| + |b| \]
Here, we can apply it to \( x \) and \( iy \):
\[ |z| = |x + iy| \leq |x| + |iy| = |x| + |y| \]

3. **Square both sides**:
To analyze the relationship, we can square both sides of the inequality:
\[ |z|^2 = x^2 + y^2 \]
and
\[ (|x| + |y|)^2 = |x|^2 + |y|^2 + 2|x||y| \]

4. **Using the Cauchy-Schwarz inequality**:
According to the Cauchy-Schwarz inequality:
\[ (|x| + |y|)^2 \leq 2(|x|^2 + |y|^2) \]
Thus, we have:
\[ |x|^2 + |y|^2 \geq 2|x||y| \]

5. **Combine the inequalities**:
From the previous steps, we have:
\[ |x| + |y| \leq \sqrt{2(x^2 + y^2)} = \sqrt{2}|z| \]

6. **Identify \( k \)**:
From the inequality \( |x| + |y| \leq \sqrt{2}|z| \), we can see that \( k = \sqrt{2} \).

### Final Result:
Thus, the value of \( k \) is:
\[ \boxed{\sqrt{2}} \]