If `| z _(1)| = | z _(2 )| and arg ((z _(1))/( z_(2)) ) = pi,` then `z _(1) + z _(2)` is equal to

Options

A. 0
B. purely imagninary
C. purely real
D. None of these

Answer

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

1. **Given**: \( |z_1| = |z_2| \) and \( \arg\left(\frac{z_1}{z_2}\right) = \pi \).

2. **Understanding the Modulus Condition**:
Since \( |z_1| = |z_2| \), we can denote both moduli as \( r \). Therefore, we can write:
\[ z_1 = r(\cos \theta_1 + i \sin \theta_1) \quad \text{and} \quad z_2 = r(\cos \theta_2 + i \sin \theta_2) \]
where \( \theta_1 = \arg(z_1) \) and \( \theta_2 = \arg(z_2) \).

3. **Understanding the Argument Condition**:
The condition \( \arg\left(\frac{z_1}{z_2}\right) = \pi \) implies:
\[ \arg(z_1) - \arg(z_2) = \pi \]
This means:
\[ \theta_1 - \theta_2 = \pi \quad \Rightarrow \quad \theta_2 = \theta_1 - \pi \]

4. **Substituting for \( z_2 \)**:
Now substituting \( \theta_2 \) into the expression for \( z_2 \):
\[ z_2 = r(\cos(\theta_1 - \pi) + i \sin(\theta_1 - \pi) \]
Using the properties of trigonometric functions, we know:
\[ \cos(\theta_1 - \pi) = -\cos(\theta_1) \quad \text{and} \quad \sin(\theta_1 - \pi) = -\sin(\theta_1) \]
Therefore:
\[ z_2 = r(-\cos(\theta_1) - i \sin(\theta_1) ) = -r(\cos(\theta_1) + i \sin(\theta_1)) = -z_1 \]

5. **Finding \( z_1 + z_2 \)**:
Now we can find \( z_1 + z_2 \):
\[ z_1 + z_2 = z_1 + (-z_1) = 0 \]

Thus, the final result is:
\[ z_1 + z_2 = 0 \]