If `|z_1|=|z_2|` and `arg((z_1)/(z_2))=pi`, then `z_1+z_2` is equal to (a) 0 (b) purely imaginary (c) purely real (d) none of these

Options

A. 0
B. purely imaginary
C. purely real
D. none of these

Explanation

To solve the problem, we start with the given conditions:

1. \( |z_1| = |z_2| \)
2. \( \arg\left(\frac{z_1}{z_2}\right) = \pi \)

### Step 1: Understand the implications of the given conditions
Since \( |z_1| = |z_2| \), we can denote this common magnitude as \( R \). Therefore, we can write:
\[ |z_1| = R \quad \text{and} \quad |z_2| = R \]

### Step 2: Analyze the argument condition
The condition \( \arg\left(\frac{z_1}{z_2}\right) = \pi \) implies:
\[ \arg(z_1) - \arg(z_2) = \pi \]
Let \( \arg(z_1) = \theta_1 \) and \( \arg(z_2) = \theta_2 \). Then:
\[ \theta_1 - \theta_2 = \pi \]
This means:
\[ \theta_1 = \theta_2 + \pi \]

### Step 3: Express \( z_1 \) and \( z_2 \) in exponential form
Using Euler's formula, we can express \( z_1 \) and \( z_2 \) as:
\[ z_1 = R e^{i\theta_1} = R e^{i(\theta_2 + \pi)} \]
\[ z_2 = R e^{i\theta_2} \]

### Step 4: Substitute the expressions for \( z_1 \) and \( z_2 \)
Substituting \( \theta_1 \) into the expression for \( z_1 \):
\[ z_1 = R e^{i\theta_2} e^{i\pi} = R e^{i\theta_2} (-1) = -R e^{i\theta_2} \]
Thus, we have:
\[ z_1 = -z_2 \]

### Step 5: Find \( z_1 + z_2 \)
Now, we can add \( z_1 \) and \( z_2 \):
\[ z_1 + z_2 = -z_2 + z_2 = 0 \]

### Conclusion
Thus, we conclude that:
\[ z_1 + z_2 = 0 \]

### Final Answer
The answer is (a) 0.

---