If `z _(1) and z _(2)` are two complex numbers such that `|z _(1)| = |z _(2)| `and arg `(z_(1)) + ` arg `(z _(2)) = pi,` then `z _(1)` is equal to

Options

A. `2 vec z _(2)`
B. `vecz _(2)`
C. `- vec z _(2)`
D. None of these

Answer

To solve the problem, we need to analyze the given conditions about the complex numbers \( z_1 \) and \( z_2 \).

### Step 1: Understand the given conditions
We have:
1. \( |z_1| = |z_2| \) (the magnitudes of the two complex numbers are equal)
2. \( \arg(z_1) + \arg(z_2) = \pi \) (the sum of their arguments is equal to \( \pi \))

### Step 2: Represent the complex numbers
We can represent the complex numbers in polar form:
- Let \( z_1 = r_1 (\cos \theta_1 + i \sin \theta_1) \)
- Let \( z_2 = r_2 (\cos \theta_2 + i \sin \theta_2) \)

From the first condition, since \( |z_1| = |z_2| \), we have \( r_1 = r_2 = r \).

### Step 3: Use the argument condition
From the second condition, we know:
\[ \arg(z_1) + \arg(z_2) = \pi \implies \theta_1 + \theta_2 = \pi \]
This implies:
\[ \theta_2 = \pi - \theta_1 \]

### Step 4: Substitute \( \theta_2 \) into the expression for \( z_2 \)
Now we can express \( z_2 \) in terms of \( z_1 \):
\[ z_2 = r (\cos(\pi - \theta_1) + i \sin(\pi - \theta_1)) \]
Using the trigonometric identities:
- \( \cos(\pi - \theta) = -\cos(\theta) \)
- \( \sin(\pi - \theta) = \sin(\theta) \)

We can rewrite \( z_2 \):
\[ z_2 = r (-\cos \theta_1 + i \sin \theta_1) = -r (\cos \theta_1 - i \sin \theta_1) \]

### Step 5: Relate \( z_2 \) to \( z_1 \)
Since \( z_1 = r (\cos \theta_1 + i \sin \theta_1) \), we can express \( z_2 \) as:
\[ z_2 = -z_1^* \]
where \( z_1^* \) is the conjugate of \( z_1 \).

### Conclusion
Thus, we find that:
\[ z_2 = -z_1^* \]

### Final Answer
Therefore, \( z_1 \) is equal to:
\[ z_1 = -z_2^* \]