The value of `lambda` for which the inequality `|z_(1)/|z_(1)|+z_(2)/|z_(2)|| le lambda` is true for all `z_(1),z_(2) in C`, is

Options

A. 1
B. 2
C. 3
D. none of these

Explanation

To solve the inequality \( \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq \lambda \) for all \( z_1, z_2 \in \mathbb{C} \), we will follow these steps:

### Step 1: Understand the Components
The terms \( \frac{z_1}{|z_1|} \) and \( \frac{z_2}{|z_2|} \) represent the unit vectors in the direction of \( z_1 \) and \( z_2 \), respectively. The modulus of a complex number \( z \) is defined as \( |z| \).

### Step 2: Use the Triangle Inequality
We will apply the triangle inequality, which states that for any two complex numbers \( a \) and \( b \):
\[ |a + b| \leq |a| + |b| \]
In our case, let \( a = \frac{z_1}{|z_1|} \) and \( b = \frac{z_2}{|z_2|} \).

### Step 3: Apply the Triangle Inequality
Using the triangle inequality, we have:
\[ \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq \left| \frac{z_1}{|z_1|} \right| + \left| \frac{z_2}{|z_2|} \right| \]

### Step 4: Simplify the Right-Hand Side
Since \( \left| \frac{z_1}{|z_1|} \right| = 1 \) and \( \left| \frac{z_2}{|z_2|} \right| = 1 \) (as they are unit vectors), we can simplify the right-hand side:
\[ \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 1 + 1 = 2 \]

### Step 5: Conclude the Value of \( \lambda \)
From the inequality we derived, we see that:
\[ \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 2 \]
This means that for the original inequality \( \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq \lambda \) to hold for all \( z_1, z_2 \in \mathbb{C} \), the maximum value of \( \lambda \) must be 2.

Thus, the value of \( \lambda \) is:
\[ \lambda = 2 \]

### Final Answer
The value of \( \lambda \) for which the inequality holds for all \( z_1, z_2 \in \mathbb{C} \) is \( \lambda = 2 \).

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