If `z_(1),z_(2)` are any two complex numbers such that `|z_(1)+z_(2)|=|z_(1)|+z_(2)|,` which one of the following is correct ?

Options

A. `z_(1)=alphaz_(2)with alphainR`
B. `z_(1) ge 0or z_(2)ge0`
C. `z_(1)=alphaz_(2)with alphagt0`
D. `|z_(1)|=|z_(2)|`

Explanation

To solve the problem, we need to analyze the condition given for the complex numbers \( z_1 \) and \( z_2 \):

Given:
\[ |z_1 + z_2| = |z_1| + |z_2| \]

This condition implies that \( z_1 \) and \( z_2 \) are in the same direction in the complex plane. Let's break down the solution step by step.

### Step 1: Understanding the Condition
The equality \( |z_1 + z_2| = |z_1| + |z_2| \) holds true if and only if \( z_1 \) and \( z_2 \) are non-negative multiples of each other. This is a property of the triangle inequality in complex numbers.

### Step 2: Expressing Complex Numbers
Let:
\[ z_1 = r_1 e^{i\theta} \quad \text{and} \quad z_2 = r_2 e^{i\phi} \]
where \( r_1 = |z_1| \), \( r_2 = |z_2| \), \( \theta \) is the argument of \( z_1 \), and \( \phi \) is the argument of \( z_2 \).

### Step 3: Applying the Condition
Using the polar form, we can express the left-hand side:
\[ |z_1 + z_2| = |r_1 e^{i\theta} + r_2 e^{i\phi}| \]
Using the properties of magnitudes:
\[ = |r_1 + r_2 e^{i(\phi - \theta)}| \]

### Step 4: Squaring Both Sides
Now, we square both sides of the original condition:
\[ |z_1 + z_2|^2 = (|z_1| + |z_2|)^2 \]
This expands to:
\[ |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(\phi - \theta) = |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \]

### Step 5: Simplifying the Equation
Canceling \( |z_1|^2 + |z_2|^2 \) from both sides, we get:
\[ 2|z_1||z_2| \cos(\phi - \theta) = 2|z_1||z_2| \]
Dividing both sides by \( 2|z_1||z_2| \) (assuming \( |z_1| \) and \( |z_2| \) are not zero), we obtain:
\[ \cos(\phi - \theta) = 1 \]

### Step 6: Conclusion
The equation \( \cos(\phi - \theta) = 1 \) implies that:
\[ \phi - \theta = 0 \quad \Rightarrow \quad \theta = \phi \]
This means that the arguments of \( z_1 \) and \( z_2 \) are equal, which leads us to conclude that:
\[ z_1 = \alpha z_2 \quad \text{for some real number } \alpha \]

### Final Answer
Thus, the correct statement is:
\[ z_1 = \alpha z_2 \quad \text{where } \alpha \text{ is a real number.} \]